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23m^2+11m=0
a = 23; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·23·0
Δ = 121
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{121}=11$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*23}=\frac{-22}{46} =-11/23 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*23}=\frac{0}{46} =0 $
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