s^2+18s+18=0

Solution for s^2+18s+18=0 equation:

s^2+18s+18=0

a = 1; b = 18; c = +18;
Δ = b2-4ac
Δ = 182-4·1·18
Δ = 252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{7}}{2*1}=\frac{-18-6\sqrt{7}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{7}}{2*1}=\frac{-18+6\sqrt{7}}{2}
$