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2x(3x+2)=1
We move all terms to the left:
2x(3x+2)-(1)=0
We multiply parentheses
6x^2+4x-1=0
a = 6; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·6·(-1)
Δ = 40
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*6}=\frac{-4-2\sqrt{10}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*6}=\frac{-4+2\sqrt{10}}{12} $
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