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3-4x=3x^2-6x+3
We move all terms to the left:
3-4x-(3x^2-6x+3)=0
We get rid of parentheses
-3x^2-4x+6x-3+3=0
We add all the numbers together, and all the variables
-3x^2+2x=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:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-3}=\frac{-4}{-6} =2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-3}=\frac{0}{-6} =0 $
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