n^2-14n+20=71

Solution for n^2-14n+20=71 equation:

n^2-14n+20=71

We move all terms to the left:

n^2-14n+20-(71)=0

We add all the numbers together, and all the variables

n^2-14n-51=0

a = 1; b = -14; c = -51;
Δ = b2-4ac
Δ = -142-4·1·(-51)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-20}{2*1}=\frac{-6}{2}
=-3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+20}{2*1}=\frac{34}{2}
=17
$