n^2-18n+52=4

Solution for n^2-18n+52=4 equation:

n^2-18n+52=4

We move all terms to the left:

n^2-18n+52-(4)=0

We add all the numbers together, and all the variables

n^2-18n+48=0

a = 1; b = -18; c = +48;
Δ = b2-4ac
Δ = -182-4·1·48
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{33}}{2*1}=\frac{18-2\sqrt{33}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{33}}{2*1}=\frac{18+2\sqrt{33}}{2}
$