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3x^2+47x-1008=0
a = 3; b = 47; c = -1008;
Δ = b2-4ac
Δ = 472-4·3·(-1008)
Δ = 14305
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-\sqrt{14305}}{2*3}=\frac{-47-\sqrt{14305}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+\sqrt{14305}}{2*3}=\frac{-47+\sqrt{14305}}{6} $
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