r2-13r+36=0

Solution for r2-13r+36=0 equation:

r2-13r+36=0

We add all the numbers together, and all the variables

r^2-13r+36=0

a = 1; b = -13; c = +36;
Δ = b2-4ac
Δ = -132-4·1·36
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*1}=\frac{8}{2}
=4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*1}=\frac{18}{2}
=9
$