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(3x^2)=96
We move all terms to the left:
(3x^2)-(96)=0
a = 3; b = 0; c = -96;
Δ = b2-4ac
Δ = 02-4·3·(-96)
Δ = 1152
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{2}}{2*3}=\frac{0-24\sqrt{2}}{6} =-\frac{24\sqrt{2}}{6} =-4\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{2}}{2*3}=\frac{0+24\sqrt{2}}{6} =\frac{24\sqrt{2}}{6} =4\sqrt{2} $
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