r2+18r-216=0

Solution for r2+18r-216=0 equation:

r2+18r-216=0

We add all the numbers together, and all the variables

r^2+18r-216=0

a = 1; b = 18; c = -216;
Δ = b2-4ac
Δ = 182-4·1·(-216)
Δ = 1188
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1188}=\sqrt{36*33}=\sqrt{36}*\sqrt{33}=6\sqrt{33}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{33}}{2*1}=\frac{-18-6\sqrt{33}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{33}}{2*1}=\frac{-18+6\sqrt{33}}{2}
$