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3x^2-2x-40=0
a = 3; b = -2; c = -40;
Δ = b2-4ac
Δ = -22-4·3·(-40)
Δ = 484
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-22}{2*3}=\frac{-20}{6} =-3+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+22}{2*3}=\frac{24}{6} =4 $
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