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3(2x^2-4)=12
We move all terms to the left:
3(2x^2-4)-(12)=0
We multiply parentheses
6x^2-12-12=0
We add all the numbers together, and all the variables
6x^2-24=0
a = 6; b = 0; c = -24;
Δ = b2-4ac
Δ = 02-4·6·(-24)
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*6}=\frac{-24}{12} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*6}=\frac{24}{12} =2 $
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