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10m+20m^2=0
a = 20; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·20·0
Δ = 100
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{100}=10$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*20}=\frac{-20}{40} =-1/2 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*20}=\frac{0}{40} =0 $
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