2x^2+7=x^2+8x

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

2x^2+7=x^2+8x

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-8x+7=0

a = 1; b = -8; c = +7;
Δ = b2-4ac
Δ = -82-4·1·7
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-6}{2*1}=\frac{2}{2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+6}{2*1}=\frac{14}{2}
=7
$