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2p^2+p-120=0
a = 2; b = 1; c = -120;
Δ = b2-4ac
Δ = 12-4·2·(-120)
Δ = 961
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{961}=31$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*2}=\frac{-32}{4} =-8 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*2}=\frac{30}{4} =7+1/2 $
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