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15b^2+21b=0
a = 15; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·15·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*15}=\frac{-42}{30} =-1+2/5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*15}=\frac{0}{30} =0 $
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