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6x^2+14x-96=0
a = 6; b = 14; c = -96;
Δ = b2-4ac
Δ = 142-4·6·(-96)
Δ = 2500
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{2500}=50$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-50}{2*6}=\frac{-64}{12} =-5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+50}{2*6}=\frac{36}{12} =3 $
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