2r^2+18r+27=0

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Solution for 2r^2+18r+27=0 equation:



2r^2+18r+27=0
a = 2; b = 18; c = +27;
Δ = b2-4ac
Δ = 182-4·2·27
Δ = 108
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:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{3}}{2*2}=\frac{-18-6\sqrt{3}}{4} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{3}}{2*2}=\frac{-18+6\sqrt{3}}{4} $

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