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15m-25m^2=0
a = -25; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·(-25)·0
Δ = 225
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{225}=15$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*-25}=\frac{-30}{-50} =3/5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*-25}=\frac{0}{-50} =0 $
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