n^2=4n+40

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

n^2=4n+40

We move all terms to the left:

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

We get rid of parentheses

n^2-4n-40=0

a = 1; b = -4; c = -40;
Δ = b2-4ac
Δ = -42-4·1·(-40)
Δ = 176
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{11}}{2*1}=\frac{4-4\sqrt{11}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{11}}{2*1}=\frac{4+4\sqrt{11}}{2}
$