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96=3x^2+4x
We move all terms to the left:
96-(3x^2+4x)=0
We get rid of parentheses
-3x^2-4x+96=0
a = -3; b = -4; c = +96;
Δ = b2-4ac
Δ = -42-4·(-3)·96
Δ = 1168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{73}}{2*-3}=\frac{4-4\sqrt{73}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{73}}{2*-3}=\frac{4+4\sqrt{73}}{-6} $
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