# How do you solve 20x^2 = 13 - x?

Apr 7, 2018

$x \approx 0.78 , 0.83$

#### Explanation:

Rearranging the equation
$\implies 20 {x}^{2} + x - 13 = 0 \implies \left(1\right)$
$a {x}^{2} + b x + c = 0$$\implies$standard equation

Comparing ($1$) with standard equation

$a = 20 , b = 1 , c = - 13$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

=>(-1+-sqrt(1^2-4(20×-13)))/(2(20))

$\implies \frac{- 1 \pm \sqrt{1 - 4 \left(- 260\right)}}{40}$

$\implies \frac{- 1 \pm \sqrt{1 + 1040}}{40}$

$\implies \frac{- 1 \pm \sqrt{1041}}{40}$

$\implies \frac{- 1 \pm 32.26}{40}$

$x \approx \frac{- 1 + 32.26}{40} , \frac{- 1 - 32.26}{40}$

$x \approx \frac{31.26}{40} , - \frac{33.26}{40}$

$x \approx 0.78 , 0.83$