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20m^2+48m=0
a = 20; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·20·0
Δ = 2304
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{2304}=48$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*20}=\frac{-96}{40} =-2+2/5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*20}=\frac{0}{40} =0 $
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