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6x(3x+4)=96
We move all terms to the left:
6x(3x+4)-(96)=0
We multiply parentheses
18x^2+24x-96=0
a = 18; b = 24; c = -96;
Δ = b2-4ac
Δ = 242-4·18·(-96)
Δ = 7488
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7488}=\sqrt{576*13}=\sqrt{576}*\sqrt{13}=24\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24\sqrt{13}}{2*18}=\frac{-24-24\sqrt{13}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24\sqrt{13}}{2*18}=\frac{-24+24\sqrt{13}}{36} $
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