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m^2-5m-22=0
a = 1; b = -5; c = -22;
Δ = b2-4ac
Δ = -52-4·1·(-22)
Δ = 113
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}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{113}}{2*1}=\frac{5-\sqrt{113}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{113}}{2*1}=\frac{5+\sqrt{113}}{2} $
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