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3x^2+13x-47=0
a = 3; b = 13; c = -47;
Δ = b2-4ac
Δ = 132-4·3·(-47)
Δ = 733
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{-(13)-\sqrt{733}}{2*3}=\frac{-13-\sqrt{733}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{733}}{2*3}=\frac{-13+\sqrt{733}}{6} $
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