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(2x+1)(3x-2)=2
We move all terms to the left:
(2x+1)(3x-2)-(2)=0
We multiply parentheses ..
(+6x^2-4x+3x-2)-2=0
We get rid of parentheses
6x^2-4x+3x-2-2=0
We add all the numbers together, and all the variables
6x^2-1x-4=0
a = 6; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·6·(-4)
Δ = 97
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{-(-1)-\sqrt{97}}{2*6}=\frac{1-\sqrt{97}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{97}}{2*6}=\frac{1+\sqrt{97}}{12} $
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