The Quadratic Formula

There are a few methods for solving quadratic equations. If the quadratic equation is not easily solvable by the factoring method, we resort to using either completing the square or the quadratic formula.

So, the quadratic formula is a guaranteed or surefire way of solving quadratic equations. That means all quadratic equations can be solved by the quadratic formula.

With that said, if we are given a quadratic equation of the form


where a, b, and c are real numbers but a does not equal to zero a \ne 0; and x is the unknown variable, we simply take the values of a, b, and c, plug them into the quadratic formula, then simplify to find the answers. The answers to the quadratic equations are called solutions, zeros, or roots.

But before we can apply the quadratic formula, we need to make sure that the quadratic equation is in the standard form. A quadratic equation is expressed in standard form if all the variables and coefficients are found on one side of the equation, and the opposite across the equal symbol is just zero. That is, if possible, we rewrite and rearrange any equation into the form \color{red}ax^2+bx+c=0. The word quadratic came from the Latin word “quadratus” which means “square”. In algebra, the word square denotes multiplying a number or a variable by itself. Notice that the highest exponent of the variable x in a quadratic equation is 2 (x raised to the 2nd power).

x=(-b+ or - sqrt(b^2-4ac))/2a

Notice also in the quadratic formula above, there is a plus or minus symbol \color{red}\large{ \pm} within the formula which implies that we are going to consider two cases when solving for the solutions.

Discriminant of the Quadratic Formula

The discriminant is the part of the quadratic formula that is inside the square root. For emphasis, it is just the expression \color{red}{b^2-4ac} excluding the radical symbol.

\LARGE{x = {{ - b \,\pm\, \sqrt {{\color{red}{b^2} - 4ac}} } \over {2a}}}

The discriminant of the quadratic formula provides very useful information regarding the nature of the solutions. In fact, it can determine the number and type of the solutions of a quadratic equation.

In addition, when dealing with discriminant we don’t really care how big or small the number is. Instead, we are just interested if it is positive, zero, or negative. 

  • If the discriminant is positive \color{red}\large{+}, then the quadratic equation has two distinct real solutions.
  • If the discriminant is zero \color{red}\large{0}, then the quadratic equation has exactly one real solution.
  • If the discriminant is negative \color{red}\large{-}, then the quadratic equation has two complex solutions, therefore no real solution
if b^2-4ac>0 then there are 2 real roots. if b^2-4ac=0 then there is 1 real root. of b^2-4ac <0 then there are 2 complex roots.

Examples of Using the Quadratic Formula

Example 1: Solve {x^2} + 4x - 12 = 0 using the Quadratic Formula.

This equation can easily be solved by factoring method. But for the sake of this lesson, we are asked to solve it using the quadratic formula.

The equation is already in the standard form. The values of a, b, and c can be readily identified.

in x^2+4x-12=0 a=1, b=4, and c=-12

Plug them into the formula then simplify.

x sub 1 equals 2, x sub 2 equals -6

The solutions are 2 and -6.

Let’s graph f\left( x \right) = {x^2} + 4x - 12 and see what we get.

graph of f(x)=x^2+4x-12 where the x-intercepts are -6 and 2

As you can see, the graph crosses or intersects the x-axis at -6 and 2. The x-intercepts of the parabola match with the solutions that we have computed using the quadratic formula.

Example 2: Solve 3{x^2} - 17x + 3 = - 7 using the Quadratic Formula.

The right-hand side of the equation is not completely zero. There is -7 hanging in there. We can get rid of it by adding both sides of the equation by 7.

3{x^2} - 17x + 3 {\color{red}+ 7} = - 7 {\color{red}+ 7}

3{x^2} - 17x + 10 = 0

Clearly, the quadratic equation is now in the standard form. Before we solve it, let’s check out its discriminant to see what kind of solutions we are going to get.

From 3{x^2} - 17x + 10 = 0

a = 3

b = -17

c = 10

The discriminant is b^2-4ac, thus

{b^2} - 4ac = {\left( { - 17} \right)^2} - 4\left( 3 \right)\left( {10} \right)

= 289 - 120

= 169

Since the discriminant is positive, we will have 2 real solutions or roots.

Let’s proceed with solving the quadratic equation. Identify the a, b, and c of the equation.

from 3x^2-17x+10, a=3, b=-17 and c=10

Substitute the values into the quadratic formula then simplify.

x sub 1 equals 5, and x sub 2 equals 2/3

The solutions are 5 and \Large{{2 \over 3}}.

The graph of f\left( x \right) = 3{x^2} - 17x + 10. Observe that the x-intercepts are the solutions to the quadratic equation.

the graph of f(x)=3x^2-17x+10 with x-intercepts at 5 and 2/3

Example 3: Use the Quadratic Formula to solve the quadratic equation 4{x^2} - x + 9 = 3x + 8.

Since either side of the equation is not zero, it means that the equation is not written in standard form. Let’s move everything to the left side by making the right side equal to zero. Subtract both sides by 8, then by 3x.

4x^2-4x+1=0 is in the standard form

From 4{x^2} - 4x + 1 = 0, we have the values a=4, b=-4, and c=1.

Calculating the discriminant, we get

{b^2} - 4ac = {\left( { - 4} \right)^2} - 4\left( 4 \right)\left( 1 \right)

= 16 - 16

= 0

Since the discriminant is equal to 0, the quadratic equation has \color{red}1 real solution.

Let’s check if indeed the quadratic equation has one real root.

in 4x^2-4x+1=0 a=4, b=-4 and c=1

Plug in the values of a, b, and c into the quadratic formula then simplify.

the only root or solution is 1/2

The solution is \Large{1 \over 2}. Yes, there is only one real number solution as predicted by the discriminant.

This is the graph of f\left( x \right) = 4{x^2} - 4x + 1. Notice, the graph does NOT cross or intersect the x-axis at \large{1 \over 2}. Instead, it only touches it.

the graph of parabola that opens upward and with x-intercept of 1/2

Example 4: Use the Quadratic Formula to solve the quadratic equation 3{x^2} - 11x + 4 = 2{x^2} - 5x.

We can transform the quadratic equation into the standard form by adding 5x to both sides of the equation then subtracting by 2x^2.

the standard form of a quadratic equation is x^2-6x+4=0

Determine the values of a, b, and c from the standard form of the quadratic equation.

in x^2-6x+4=0, a=1, b=-6, and c=4

Substitute the values into the formula then simplify.

x sub 1 equals 3+sqrt(5) and x sub 2 equals 3 - sqrt(5)

The solutions are 3 + \sqrt 5 and 3 - \sqrt 5 . Notice that the roots are irrational numbers.

Here’s the graph of f\left( x \right) = {x^2} - 6x + 4. The x-intercepts are the solutions to the given quadratic equation.

the graph of f(x)=x^2-6x+4 is a parabola that opens upward with x intercepts at 3+sqrt(5) and 3-sqrt(5)

Example 5: Solve 5{x^2} + 3x + 4 = 4{x^2} + 7x - 9 using the Quadratic Formula.

The quadratic equation is kind of a mess. We need to rewrite it in standard form. We can do that by adding both sides by 9. Next, subtract both sides by 7x. Finally, subtract both sides by 4{x^2}.

the standard form is x^2-4x+13=0

From the standard form {x^2} - 4x + 13 = 0, the values of a, b, and c are as follows.

\large{a = 1}

\large{b = -4}

\large{c = 13}

Calculating the discriminant

\large{{b^2} - 4ac = {\left( { - 4} \right)^2} - 4\left( 1 \right)\left( 13 \right)}

\large{ = 16 - 52}

\large{ = - 36}

Since the discriminant is negative, the quadratic equation will have two complex solutions.

Now we solve the equation using the formula.

in x^2-4x+13=0 the values are the following; a=1, b=-4, and c=13

Substitute then simplify.

x sub 1 equals 2 - 3 i and x sub 2 equals 2 - + 3 i

The complex solutions are 2 + 3i and 2 - 3i.

Since the solutions are complex numbers, the graph of the parabola does not intersect or touch the x-axis. It also means that the graph does not have x-intercepts as you can see in the graph below.

the graph of f(x)=x^2-4x+13 does not have x-intercepts

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