2n^2=-2n+40

Solution for 2n^2=-2n+40 equation:

2n^2=-2n+40

We move all terms to the left:

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

We get rid of parentheses

2n^2+2n-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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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