N+n2=132

Solution for N+n2=132 equation:

+N2=132

We move all terms to the left:

+N2-(132)=0

We add all the numbers together, and all the variables

N^2-132=0

a = 1; b = 0; c = -132;
Δ = b2-4ac
Δ = 02-4·1·(-132)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{33}}{2*1}=\frac{0-4\sqrt{33}}{2}
=-\frac{4\sqrt{33}}{2}
=-2\sqrt{33}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{33}}{2*1}=\frac{0+4\sqrt{33}}{2}
=\frac{4\sqrt{33}}{2}
=2\sqrt{33}
$