n^2-20n+16=-3

Solution for n^2-20n+16=-3 equation:

n^2-20n+16=-3

We move all terms to the left:

n^2-20n+16-(-3)=0

We add all the numbers together, and all the variables

n^2-20n+19=0

a = 1; b = -20; c = +19;
Δ = b2-4ac
Δ = -202-4·1·19
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-18}{2*1}=\frac{2}{2}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+18}{2*1}=\frac{38}{2}
=19
$