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2(3x^2-1)=6x+5x
We move all terms to the left:
2(3x^2-1)-(6x+5x)=0
We add all the numbers together, and all the variables
2(3x^2-1)-(+11x)=0
We multiply parentheses
6x^2-(+11x)-2=0
We get rid of parentheses
6x^2-11x-2=0
a = 6; b = -11; c = -2;
Δ = b2-4ac
Δ = -112-4·6·(-2)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-13}{2*6}=\frac{-2}{12} =-1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+13}{2*6}=\frac{24}{12} =2 $
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