n^2+13=75

Solution for n^2+13=75 equation:

n^2+13=75

We move all terms to the left:

n^2+13-(75)=0

We add all the numbers together, and all the variables

n^2-62=0

a = 1; b = 0; c = -62;
Δ = b2-4ac
Δ = 02-4·1·(-62)
Δ = 248
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:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{62}}{2*1}=\frac{0-2\sqrt{62}}{2}
=-\frac{2\sqrt{62}}{2}
=-\sqrt{62}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{62}}{2*1}=\frac{0+2\sqrt{62}}{2}
=\frac{2\sqrt{62}}{2}
=\sqrt{62}
$