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3x^2+96x+720=0
a = 3; b = 96; c = +720;
Δ = b2-4ac
Δ = 962-4·3·720
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-24}{2*3}=\frac{-120}{6} =-20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+24}{2*3}=\frac{-72}{6} =-12 $
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