n2=18n+40

Solution for n2=18n+40 equation:

n2=18n+40

We move all terms to the left:

n2-(18n+40)=0

We add all the numbers together, and all the variables

n^2-(18n+40)=0

We get rid of parentheses

n^2-18n-40=0

a = 1; b = -18; c = -40;
Δ = b2-4ac
Δ = -182-4·1·(-40)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-22}{2*1}=\frac{-4}{2}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+22}{2*1}=\frac{40}{2}
=20
$