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(2x^2-1)3-(x+3)3=0
We multiply parentheses
6x^2-3x-3-9=0
We add all the numbers together, and all the variables
6x^2-3x-12=0
a = 6; b = -3; c = -12;
Δ = b2-4ac
Δ = -32-4·6·(-12)
Δ = 297
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{33}}{2*6}=\frac{3-3\sqrt{33}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{33}}{2*6}=\frac{3+3\sqrt{33}}{12} $
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