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(2x-1)(3x=1)
We move all terms to the left:
(2x-1)(3x-(1))=0
We multiply parentheses ..
(+6x^2-2x-3x+1)=0
We get rid of parentheses
6x^2-2x-3x+1=0
We add all the numbers together, and all the variables
6x^2-5x+1=0
a = 6; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·6·1
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*6}=\frac{4}{12} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*6}=\frac{6}{12} =1/2 $
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