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(3x-4)(x-2)=3x
We move all terms to the left:
(3x-4)(x-2)-(3x)=0
We add all the numbers together, and all the variables
-3x+(3x-4)(x-2)=0
We multiply parentheses ..
(+3x^2-6x-4x+8)-3x=0
We get rid of parentheses
3x^2-6x-4x-3x+8=0
We add all the numbers together, and all the variables
3x^2-13x+8=0
a = 3; b = -13; c = +8;
Δ = b2-4ac
Δ = -132-4·3·8
Δ = 73
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{73}}{2*3}=\frac{13-\sqrt{73}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{73}}{2*3}=\frac{13+\sqrt{73}}{6} $
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