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m^2-110+m=0
a = 1; b = 1; c = -110;
Δ = b2-4ac
Δ = 12-4·1·(-110)
Δ = 441
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{441}=21$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-21}{2*1}=\frac{-22}{2} =-11 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+21}{2*1}=\frac{20}{2} =10 $
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