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

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

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

We move all terms to the left:

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

We multiply parentheses

2x^2-(x^2+8x)-20=0

We get rid of parentheses

2x^2-x^2-8x-20=0

We add all the numbers together, and all the variables

x^2-8x-20=0

a = 1; b = -8; c = -20;
Δ = b2-4ac
Δ = -82-4·1·(-20)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-12}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+12}{2*1}=\frac{20}{2}
=10
$