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2x+6x^2=96
We move all terms to the left:
2x+6x^2-(96)=0
a = 6; b = 2; c = -96;
Δ = b2-4ac
Δ = 22-4·6·(-96)
Δ = 2308
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2308}=\sqrt{4*577}=\sqrt{4}*\sqrt{577}=2\sqrt{577}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{577}}{2*6}=\frac{-2-2\sqrt{577}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{577}}{2*6}=\frac{-2+2\sqrt{577}}{12} $
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