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m^2-10m-56=0
a = 1; b = -10; c = -56;
Δ = b2-4ac
Δ = -102-4·1·(-56)
Δ = 324
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{324}=18$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-18}{2*1}=\frac{-8}{2} =-4 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+18}{2*1}=\frac{28}{2} =14 $
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