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m^2-23m+132=0
a = 1; b = -23; c = +132;
Δ = b2-4ac
Δ = -232-4·1·132
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-1}{2*1}=\frac{22}{2} =11 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+1}{2*1}=\frac{24}{2} =12 $
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