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15y^2+26y=0
a = 15; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·15·0
Δ = 676
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{676}=26$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*15}=\frac{-52}{30} =-1+11/15 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*15}=\frac{0}{30} =0 $
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