n^2=n+20

Solution for n^2=n+20 equation:

n^2=n+20

We move all terms to the left:

n^2-(n+20)=0

We get rid of parentheses

n^2-n-20=0

We add all the numbers together, and all the variables

n^2-1n-20=0

a = 1; b = -1; c = -20;
Δ = b2-4ac
Δ = -12-4·1·(-20)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*1}=\frac{-8}{2}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*1}=\frac{10}{2}
=5
$