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18x^2+40x=0
a = 18; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·18·0
Δ = 1600
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{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*18}=\frac{-80}{36} =-2+2/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*18}=\frac{0}{36} =0 $
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