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3x^2+68x-18=0
a = 3; b = 68; c = -18;
Δ = b2-4ac
Δ = 682-4·3·(-18)
Δ = 4840
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4840}=\sqrt{484*10}=\sqrt{484}*\sqrt{10}=22\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-22\sqrt{10}}{2*3}=\frac{-68-22\sqrt{10}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+22\sqrt{10}}{2*3}=\frac{-68+22\sqrt{10}}{6} $
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