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m2+m=20
We move all terms to the left:
m2+m-(20)=0
We add all the numbers together, and all the variables
m^2+m-20=0
a = 1; b = 1; c = -20;
Δ = b2-4ac
Δ = 12-4·1·(-20)
Δ = 81
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{81}=9$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*1}=\frac{-10}{2} =-5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*1}=\frac{8}{2} =4 $
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