How do you solve 4x^2-13x=12?

Jun 27, 2015

First subtract $12$ from both sides to get $4 {x}^{2} - 13 x - 12 = 0$, then use the quadratic formula to find:

$x = \frac{13 \pm 19}{8}$

That is $x = 4$ or $x = - \frac{3}{4}$

Explanation:

$4 {x}^{2} - 13 x - 12$ is in the form $a {x}^{2} + b x + c$ with $a = 4$, $b = - 13$ and $c = - 12$

It discriminant is given by the formula:

$\Delta = {b}^{2} - 4 a c = {13}^{2} - \left(4 \times 4 \times - 12\right) = 169 + 192 = 361 = {19}^{2}$

Since $\Delta$ is positive and a perfect square, the roots of our quadratic are rational, given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{13 \pm 19}{8}$

That is $x = \frac{32}{8} = 4$ or $x = - \frac{6}{8} = - \frac{3}{4}$