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m^2-13m-14=0
a = 1; b = -13; c = -14;
Δ = b2-4ac
Δ = -132-4·1·(-14)
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-15}{2*1}=\frac{-2}{2} =-1 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+15}{2*1}=\frac{28}{2} =14 $
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