r^2-18r+32=0

Solution for r^2-18r+32=0 equation:

r^2-18r+32=0

a = 1; b = -18; c = +32;
Δ = b2-4ac
Δ = -182-4·1·32
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-14}{2*1}=\frac{4}{2}
=2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+14}{2*1}=\frac{32}{2}
=16
$