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m2+5m=2m^2
We move all terms to the left:
m2+5m-(2m^2)=0
determiningTheFunctionDomain -2m^2+m2+5m=0
We add all the numbers together, and all the variables
-1m^2+5m=0
a = -1; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-1)·0
Δ = 25
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{25}=5$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-1}=\frac{-10}{-2} =+5 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-1}=\frac{0}{-2} =0 $
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