n^2-32n+63=0

Solution for n^2-32n+63=0 equation:

n^2-32n+63=0

a = 1; b = -32; c = +63;
Δ = b2-4ac
Δ = -322-4·1·63
Δ = 772
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{772}=\sqrt{4*193}=\sqrt{4}*\sqrt{193}=2\sqrt{193}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{193}}{2*1}=\frac{32-2\sqrt{193}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{193}}{2*1}=\frac{32+2\sqrt{193}}{2}
$