# How do you solve 2x^2-5x=4?

May 30, 2015

$2 {x}^{2} - 5 x = 4$
is equivalent to
$\textcolor{w h i t e}{\text{XXXXX}}$$2 {x}^{2} - 5 x - 4 = 0$

For a quadratic in the general form
$\textcolor{w h i t e}{\text{XXXXX}}$$a {x}^{2} + b x + c = 0$
the solutions can be obtained using the quadratic root formula
$\textcolor{w h i t e}{\text{XXXXX}}$$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case the roots are:
$\textcolor{w h i t e}{\text{XXXXX}}$$x = \frac{5 \pm \sqrt{25 + 32}}{4}$

$x = \frac{5 + \sqrt{57}}{4}$ or $x = \frac{5 - \sqrt{57}}{4}$