r^2-16r+36=0

Solution for r^2-16r+36=0 equation:

r^2-16r+36=0

a = 1; b = -16; c = +36;
Δ = b2-4ac
Δ = -162-4·1·36
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{7}}{2*1}=\frac{16-4\sqrt{7}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{7}}{2*1}=\frac{16+4\sqrt{7}}{2}
$