2x^2+8x=x^2+4x

Solution for 2x^2+8x=x^2+4x equation:

2x^2+8x=x^2+4x

We move all terms to the left:

2x^2+8x-(x^2+4x)=0

We get rid of parentheses

2x^2-x^2+8x-4x=0

We add all the numbers together, and all the variables

x^2+4x=0

a = 1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·1·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*1}=\frac{0}{2}
=0
$