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12m^2+25m=0
a = 12; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·12·0
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*12}=\frac{-50}{24} =-2+1/12 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*12}=\frac{0}{24} =0 $
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