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3m+2m^2=m(m-2)
We move all terms to the left:
3m+2m^2-(m(m-2))=0
We calculate terms in parentheses: -(m(m-2)), so:We get rid of parentheses
m(m-2)
We multiply parentheses
m^2-2m
Back to the equation:
-(m^2-2m)
2m^2-m^2+3m+2m=0
We add all the numbers together, and all the variables
m^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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