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3x^2+2x=125
We move all terms to the left:
3x^2+2x-(125)=0
a = 3; b = 2; c = -125;
Δ = b2-4ac
Δ = 22-4·3·(-125)
Δ = 1504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1504}=\sqrt{16*94}=\sqrt{16}*\sqrt{94}=4\sqrt{94}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{94}}{2*3}=\frac{-2-4\sqrt{94}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{94}}{2*3}=\frac{-2+4\sqrt{94}}{6} $
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