n2=18n+4

Solution for n2=18n+4 equation:

n2=18n+4

We move all terms to the left:

n2-(18n+4)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

n^2-18n-4=0

a = 1; b = -18; c = -4;
Δ = b2-4ac
Δ = -182-4·1·(-4)
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{85}}{2*1}=\frac{18-2\sqrt{85}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{85}}{2*1}=\frac{18+2\sqrt{85}}{2}
$