n^2+n-20=52

Solution for n^2+n-20=52 equation:

n^2+n-20=52

We move all terms to the left:

n^2+n-20-(52)=0

We add all the numbers together, and all the variables

n^2+n-72=0

a = 1; b = 1; c = -72;
Δ = b2-4ac
Δ = 12-4·1·(-72)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*1}=\frac{-18}{2}
=-9
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*1}=\frac{16}{2}
=8
$