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6x^2+4x=48
We move all terms to the left:
6x^2+4x-(48)=0
a = 6; b = 4; c = -48;
Δ = b2-4ac
Δ = 42-4·6·(-48)
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*6}=\frac{-4-4\sqrt{73}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{73}}{2*6}=\frac{-4+4\sqrt{73}}{12} $
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