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3x^2+14x-24=0
a = 3; b = 14; c = -24;
Δ = b2-4ac
Δ = 142-4·3·(-24)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-22}{2*3}=\frac{-36}{6} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+22}{2*3}=\frac{8}{6} =1+1/3 $
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