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3x^2-3x=x-1
We move all terms to the left:
3x^2-3x-(x-1)=0
We get rid of parentheses
3x^2-3x-x+1=0
We add all the numbers together, and all the variables
3x^2-4x+1=0
a = 3; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·3·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*3}=\frac{2}{6} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*3}=\frac{6}{6} =1 $
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