r2-13r+32=0

Solution for r2-13r+32=0 equation:

r2-13r+32=0

We add all the numbers together, and all the variables

r^2-13r+32=0

a = 1; b = -13; c = +32;
Δ = b2-4ac
Δ = -132-4·1·32
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{41}}{2*1}=\frac{13-\sqrt{41}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{41}}{2*1}=\frac{13+\sqrt{41}}{2}
$