2x^2=40+2x

Solution for 2x^2=40+2x equation:

2x^2=40+2x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-2x-40=0

a = 2; b = -2; c = -40;
Δ = b2-4ac
Δ = -22-4·2·(-40)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-18}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+18}{2*2}=\frac{20}{4}
=5
$