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3y^2+15y=0
a = 3; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·3·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{225}=15$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*3}=\frac{-30}{6} =-5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*3}=\frac{0}{6} =0 $
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