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2x(3x-2)=4
We move all terms to the left:
2x(3x-2)-(4)=0
We multiply parentheses
6x^2-4x-4=0
a = 6; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·6·(-4)
Δ = 112
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{7}}{2*6}=\frac{4-4\sqrt{7}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{7}}{2*6}=\frac{4+4\sqrt{7}}{12} $
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