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(m^2)+13m+12=0
a = 1; b = 13; c = +12;
Δ = b2-4ac
Δ = 132-4·1·12
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*1}=\frac{-24}{2} =-12 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*1}=\frac{-2}{2} =-1 $
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