4n^2+96=-40n

Solution for 4n^2+96=-40n equation:

4n^2+96=-40n

We move all terms to the left:

4n^2+96-(-40n)=0

We get rid of parentheses

4n^2+40n+96=0

a = 4; b = 40; c = +96;
Δ = b2-4ac
Δ = 402-4·4·96
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8}{2*4}=\frac{-48}{8}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8}{2*4}=\frac{-32}{8}
=-4
$