r^2+18.5r+36=0

Solution for r^2+18.5r+36=0 equation:

r^2+18.5r+36=0

a = 1; b = 18.5; c = +36;
Δ = b2-4ac
Δ = 18.52-4·1·36
Δ = 198.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18.5)-\sqrt{198.25}}{2*1}=\frac{-18.5-\sqrt{198.25}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18.5)+\sqrt{198.25}}{2*1}=\frac{-18.5+\sqrt{198.25}}{2}
$