r^2+11r+32=8

Solution for r^2+11r+32=8 equation:

r^2+11r+32=8

We move all terms to the left:

r^2+11r+32-(8)=0

We add all the numbers together, and all the variables

r^2+11r+24=0

a = 1; b = 11; c = +24;
Δ = b2-4ac
Δ = 112-4·1·24
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{-(11)-5}{2*1}=\frac{-16}{2}
=-8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*1}=\frac{-6}{2}
=-3
$