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3x^2+3x-68=0
a = 3; b = 3; c = -68;
Δ = b2-4ac
Δ = 32-4·3·(-68)
Δ = 825
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5\sqrt{33}}{2*3}=\frac{-3-5\sqrt{33}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5\sqrt{33}}{2*3}=\frac{-3+5\sqrt{33}}{6} $
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