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3s^2+18s+18=0
a = 3; b = 18; c = +18;
Δ = b2-4ac
Δ = 182-4·3·18
Δ = 108
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}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{3}}{2*3}=\frac{-18-6\sqrt{3}}{6} $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{3}}{2*3}=\frac{-18+6\sqrt{3}}{6} $
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