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m^2+21m+110=0
a = 1; b = 21; c = +110;
Δ = b2-4ac
Δ = 212-4·1·110
Δ = 1
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{1}=1$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-1}{2*1}=\frac{-22}{2} =-11 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+1}{2*1}=\frac{-20}{2} =-10 $
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