4=2n^2+2n

Solution for 4=2n^2+2n equation:

4=2n^2+2n

We move all terms to the left:

4-(2n^2+2n)=0

We get rid of parentheses

-2n^2-2n+4=0

a = -2; b = -2; c = +4;
Δ = b2-4ac
Δ = -22-4·(-2)·4
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*-2}=\frac{-4}{-4}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*-2}=\frac{8}{-4}
=-2
$