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(3x+1)xxx12=96
We move all terms to the left:
(3x+1)xxx12-(96)=0
We multiply parentheses
3x^2+x-96=0
a = 3; b = 1; c = -96;
Δ = b2-4ac
Δ = 12-4·3·(-96)
Δ = 1153
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{1153}}{2*3}=\frac{-1-\sqrt{1153}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1153}}{2*3}=\frac{-1+\sqrt{1153}}{6} $
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