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