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m2+m=18
We move all terms to the left:
m2+m-(18)=0
We add all the numbers together, and all the variables
m^2+m-18=0
a = 1; b = 1; c = -18;
Δ = b2-4ac
Δ = 12-4·1·(-18)
Δ = 73
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{-(1)-\sqrt{73}}{2*1}=\frac{-1-\sqrt{73}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{73}}{2*1}=\frac{-1+\sqrt{73}}{2} $
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