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3x^2-4x+2=2x^2+x-4
We move all terms to the left:
3x^2-4x+2-(2x^2+x-4)=0
We get rid of parentheses
3x^2-2x^2-4x-x+4+2=0
We add all the numbers together, and all the variables
x^2-5x+6=0
a = 1; b = -5; c = +6;
Δ = b2-4ac
Δ = -52-4·1·6
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*1}=\frac{4}{2} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*1}=\frac{6}{2} =3 $
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