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15m^2+25m=0
a = 15; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·15·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*15}=\frac{-50}{30} =-1+2/3 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*15}=\frac{0}{30} =0 $
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