9r^2+36r+32=0

Solution for 9r^2+36r+32=0 equation:

9r^2+36r+32=0

a = 9; b = 36; c = +32;
Δ = b2-4ac
Δ = 362-4·9·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12}{2*9}=\frac{-48}{18}
=-2+2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12}{2*9}=\frac{-24}{18}
=-1+1/3
$