n^2=4n+23

Solution for n^2=4n+23 equation:

n^2=4n+23

We move all terms to the left:

n^2-(4n+23)=0

We get rid of parentheses

n^2-4n-23=0

a = 1; b = -4; c = -23;
Δ = b2-4ac
Δ = -42-4·1·(-23)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{3}}{2*1}=\frac{4-6\sqrt{3}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{3}}{2*1}=\frac{4+6\sqrt{3}}{2}
$