2x^2-7x-40=-x^2-7x+8

Solution for 2x^2-7x-40=-x^2-7x+8 equation:

2x^2-7x-40=-x^2-7x+8

We move all terms to the left:

2x^2-7x-40-(-x^2-7x+8)=0

We get rid of parentheses

2x^2+x^2+7x-7x-8-40=0

We add all the numbers together, and all the variables

3x^2-48=0

a = 3; b = 0; c = -48;
Δ = b2-4ac
Δ = 02-4·3·(-48)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*3}=\frac{24}{6}
=4
$