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2m=4m-3m^2
We move all terms to the left:
2m-(4m-3m^2)=0
We get rid of parentheses
3m^2-4m+2m=0
We add all the numbers together, and all the variables
3m^2-2m=0
a = 3; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·3·0
Δ = 4
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{4}=2$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*3}=\frac{0}{6} =0 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*3}=\frac{4}{6} =2/3 $
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