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m^2=4m-3
We move all terms to the left:
m^2-(4m-3)=0
We get rid of parentheses
m^2-4m+3=0
a = 1; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·1·3
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$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*1}=\frac{2}{2} =1 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*1}=\frac{6}{2} =3 $
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