18x^2+54x=0.

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Solution for 18x^2+54x=0. equation:



18x^2+54x=0.
a = 18; b = 54; c = 0;
Δ = b2-4ac
Δ = 542-4·18·0
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-54}{2*18}=\frac{-108}{36} =-3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+54}{2*18}=\frac{0}{36} =0 $

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