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3p^2+9p-120=0
a = 3; b = 9; c = -120;
Δ = b2-4ac
Δ = 92-4·3·(-120)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1521}=39$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-39}{2*3}=\frac{-48}{6} =-8 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+39}{2*3}=\frac{30}{6} =5 $
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