2x^2+10x-8x-40=x^2+8x

Solution for 2x^2+10x-8x-40=x^2+8x equation:

2x^2+10x-8x-40=x^2+8x

We move all terms to the left:

2x^2+10x-8x-40-(x^2+8x)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-6x-40=0

a = 1; b = -6; c = -40;
Δ = b2-4ac
Δ = -62-4·1·(-40)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-14}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+14}{2*1}=\frac{20}{2}
=10
$