Below are ten (10) practice problems regarding the quadratic formula. The more you use the formula to solve quadratic equations, the more you become expert at it!

Use the illustration below as a guide. Notice that in order to apply the quadratic formula, we must transform the quadratic equation into the standard form, that is, $a{x^2} + bx + c = 0$ where $a \ne 0$.

The problems below have varying levels of difficulty. I encourage you to try them all. Believe me, they are actually easy! Good luck.

${x^2}\, – \,8x + 12 = 0$

Answer \begin{align*} &a = 1 \\ &b = – 8 \\ &c = 12 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ &= {{ – \left( { – 8} \right) \pm \sqrt {{{\left( { – 8} \right)}^2} – 4\left( 1 \right)\left( {12} \right)} } \over {2\left( 1 \right)}} \\ \\ & = {{8 \pm \sqrt {64 – 48} } \over 2} \\ \\ & = {{8 \pm \sqrt {16} } \over 2} \\ \\ & = {{8 \pm 4} \over 2} \\ \\ {x_1} &= {{8 + 4} \over 2} \\ \\ & = {{12} \over 2} \\ \\ & = \boxed{\color{red}6} \\ \\ {x_2} &= {{8 – 4} \over 2} \\ \\ & = {4 \over 2} \\ \\ & = \boxed{\color{red}2} \end{align*}

Therefore, the answers are ${x_1} = 6$ and ${x_2} = 2$.

$2{x^2}\, -\, x = 1$

Rewrite the quadratic equation in the standard form.

$2{x^2} – x – 1 = 0$

\begin{align*} & a = 2 \\ & b = – 1 \\ & c = – 1 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ & = {{ – \left( { – 1} \right) \pm \sqrt {{{\left( { – 1} \right)}^2} – 4\left( 2 \right)\left( { – 1} \right)} } \over {2\left( 2 \right)}} \\ \\ & = {{1 \pm \sqrt {1 + 8} } \over 4} \\ \\ & = {{1 \pm \sqrt 9 } \over 4} \\ \\ & = {{1 \pm 3} \over 4} \\ \\ {x_1} &= {{1 + 3} \over 4} \\ \\ & = {4 \over 4} \\ \\ & =\boxed{\color{red} 1} \\ \\ {x_2} &= {{1 – 3} \over 4} \\ \\ & = {{ – 2} \over 4} \\ \\ & = \boxed{\color{red}{{ – 1} \over 2}} \end{align*}

Therefore, the answers are ${x_1} = 1$ and ${x_2} = \large{{ – 1} \over 2}$.

$4{x^2} + 9 = – 12x$

Rewrite the quadratic equation in the standard form.

$4{x^2} + 12x + 9 = 0$

\begin{align*} & a = 4 \\ & b = 12 \\ & c = 9 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ & = {{ – \left( {12} \right) \pm \sqrt {{{\left( {12} \right)}^2} – 4\left( 4 \right)\left( 9 \right)} } \over {2\left( 4 \right)}} \\ \\ & = {{ – 12 \pm \sqrt {144 – 144} } \over 8} \\ \\ & = {{ – 12 \pm \sqrt 0 } \over 8} \\ \\ & = {{ – 12} \over 8} \\ \\ x &=\boxed{\color{red} {{ – 3} \over 2}} \end{align*}

Therefore, the solution is $x = \large{{ – 3} \over 2}$.

$5{x^2} = 7x + 6$

Convert the quadratic equation into the standard form.

$5{x^2} – 7x – 6 = 0$

\begin{align*} & a = 5 \\ & b = – 7 \\ & c = – 6 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ & = {{ – \left( { – 7} \right) \pm \sqrt {{{\left( { – 7} \right)}^2} – 4\left( 5 \right)\left( { – 6} \right)} } \over {2\left( 5 \right)}} \\ \\ & = {{7 \pm \sqrt {49 + 120} } \over {10}} \\ \\ & = {{7 \pm \sqrt {169} } \over {10}} \\ \\ & = {{7 \pm 13} \over {10}} \\ \\ {x_1} &= {{7 + 13} \over {10}} \\ \\ & = {{20} \over {10}} \\ \\ & =\boxed {\color{red} 2} \\ \\ {x_2} &= {{7 – 13} \over {10}} \\ \\ & = {{ – 6} \over {10}} \\ \\ & =\boxed{\color{red} {{ – 3} \over 5}} \end{align*}

Therefore, the answers are ${x_1} = 2$ and ${x_2} = \large{{ – 3} \over 5}$.

${x^2} -\,{ \large{1 \over 2}}x\, – \,{\large{3 \over {16}}} = 0$

Multiply the entire equation by the LCM of the denominators which is $16$. This will get rid of the denominators thereby giving us integer values for $a$, $b$, and $c$.

$16{x^2} – 8x – 3 = 0$

\begin{align*} & a = 16 \\ & b = – 8 \\ & c = – 3 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ & = {{ – \left( { – 8} \right) \pm \sqrt {{{\left( { – 8} \right)}^2} – 4\left( {16} \right)\left( { – 3} \right)} } \over {2\left( {16} \right)}} \\ \\ & = {{8 \pm \sqrt {64 + 192} } \over {32}} \\ \\ & = {{8 \pm \sqrt {256} } \over {32}} \\ \\ & = {{8 \pm 16} \over {32}} \\ \\ {x_1} &= {{8 + 16} \over {32}} \\ \\ & = {{24} \over {32}} \\ \\ & = \boxed{\color{red}{3 \over 4}} \\ \\ {x_2} &= {{8 – 16} \over {32}} \\ \\ & = {{ – 8} \over {32}} \\ \\ & =\boxed{\color{red} {{ – 1} \over 4}} \end{align*}

Therefore, the answers are $x_1=\large{3 \over 4}$ and $x_2=\large{{ – 1} \over 4}$.

${x^2} + 3x + 9 = 5x – 8$

Convert into standard form as ${x^2} – 2x + 17 = 0$.

\begin{align*} & a = 1 \\ & b = – 2 \\ & c = 17 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ & = {{ – \left( { – 2} \right) \pm \sqrt {{{\left( { – 2} \right)}^2} – 4\left( 1 \right)\left( {17} \right)} } \over {2\left( 1 \right)}} \\ \\ & = {{2 \pm \sqrt {4 – 68} } \over 2} \\ \\ & = {{2 \pm \sqrt { – 64} } \over 2} \\ \\ & = {{2 \pm \sqrt {64} \cdot \sqrt { – 1} } \over 2} \\ \\ & = {{2 \pm 8 \cdot i} \over 2} \\ \\ {x_1} &= {{2 + 8 \cdot i} \over 2} \\ \\ & = {{2 + 8 \cdot i} \over 2} \\ \\ & = {2 \over 2} + {8 \over 2}i \\ \\ & = \boxed{\color{red}1 + 4i} \\ \\ {x_2} &= {{2 – 8 \cdot i} \over 2} \\ \\ & = {2 \over 2} – {8 \over 2}i \\ \\ & = \boxed {\color{red}1 – 4i } \end{align*}

Therefore, the answers are $x_1=1 + 4i$ and $x_2=1 – 4i$.

${\left( {x – 2} \right)^2} = 4x$

Rewrite in standard form as ${x^2} – 8x + 4 = 0$.

\begin{align*} & a = 1 \\ & b = – 8 \\ & c = 4 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ & = {{ – \left( { – 8} \right) \pm \sqrt {{{\left( { – 8} \right)}^2} – 4\left( 1 \right)\left( 4 \right)} } \over {2\left( 1 \right)}} \\ \\ & = {{8 \pm \sqrt {64 – 16} } \over 2} \\ \\ & = {{8 \pm \sqrt {48} } \over 2} \\ \\ & = {{8 \pm \sqrt {16} \cdot \sqrt 3 } \over 2} \\ \\ & = {{8 \pm 4 \cdot \sqrt 3 } \over 2} \\ \\ {x_1} &= {{8 + 4 \cdot \sqrt 3 } \over 2} \\ \\ & = {8 \over 2} + {4 \over 2}\sqrt 3 \\ \\ & = \boxed {\color{red}4 + 2\sqrt 3} \\ \\ {x_2} &= {{8 – 4 \cdot \sqrt 3 } \over 2} \\ \\ & = {8 \over 2} – {4 \over 2}\sqrt 3 \\ \\ & = \boxed {\color{red}4 – 2\sqrt 3} \end{align*}

Hence, the answers are ${x_1} = 4 + 2\sqrt 3$ and ${x_2} = 4 – 2\sqrt 3$.

${\Large{{{x^2}} \over 4} – {x \over 2} }= 1$

To convert the quadratic equation into the standard form, simply multiply the entire equation by $4$ then subtract both sides by $4$.

${x^2} – 2x – 4 = 0$

\begin{align*} & a = 1 \\ & b = – 2 \\ & c = – 4 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ & = {{ – \left( { – 2} \right) \pm \sqrt {{{\left( { – 2} \right)}^2} – 4\left( 1 \right)\left( { – 4} \right)} } \over {2\left( 1 \right)}} \\ \\ & = {{2 \pm \sqrt {4 + 16} } \over 2} \\ \\ & = {{2 \pm \sqrt {20} } \over 2} \\ \\ & = {{2 \pm \sqrt 4 \cdot \sqrt 5 } \over 2} \\ \\ & = {{2 \pm 2 \cdot \sqrt 5 } \over 2} \\ \\ {x_1} &= {{2 + 2 \cdot \sqrt 5 } \over 2} \\ \\ & = {2 \over 2} + {2 \over 2}\sqrt 5 \\ \\ & = \boxed {\color{red}1 + \sqrt 5} \\ \\ {x_2} &= {{2 – 2 \cdot \sqrt 5 } \over 2} \\ \\ & = {2 \over 2} – {2 \over 2}\sqrt 5 \\ \\ & = \boxed{\color{red}1 – \sqrt 5} \end{align*}

Thus, the answers are ${x_1} = 1 + \sqrt 5$ and ${x_2} = 1 – \sqrt 5$.

${\left( {2x – 1} \right)^2} = \Large{x \over 3}$

If we carefully transform the given quadratic equation into the standard form, we get $12{x^2} – 13x + 3 = 0$.

\begin{align*} & a = 12 \\ & b = – 13 \\ & c = 3 \\ \\ x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \\ \\ & = {{ – \left( { – 13} \right) \pm \sqrt {{{\left( { – 13} \right)}^2} – 4\left( {12} \right)\left( 3 \right)} } \over {2\left( {12} \right)}} \\ \\ & = {{13 \pm \sqrt {169 – 144} } \over {24}} \\ \\ & = {{13 \pm \sqrt {25} } \over {24}} \\ \\ & = {{13 \pm 5} \over {24}} \\ \\ {x_1} &= {{13 + 5} \over {24}} \\ \\ & = {{18} \over {24}} \\ \\ & = \boxed {\color{red}{3 \over 4}} \\ \\ {x_2} &= {{13 – 5} \over {24}} \\ \\ & = {8 \over {24}} \\ \\ & =\boxed {\color{red} {1 \over 3}} \end{align*}

Therefore, the answers are $x_1={\Large{3 \over 4}}$ and $x_2={\Large{1 \over 3}}$.

$\left( {2x – 1} \right)\left( {x + 4} \right) = – {x^2} + 3x$

If we simplify the quadratic equation to convert it to the standard form, we should arrive at $3{x^2} + 4x – 4 = 0$.
Hence, the answers are $x_1={\Large{2 \over 3}}$ and $x_2=-2$.