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2p^2-4p-96=0
a = 2; b = -4; c = -96;
Δ = b2-4ac
Δ = -42-4·2·(-96)
Δ = 784
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{784}=28$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-28}{2*2}=\frac{-24}{4} =-6 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+28}{2*2}=\frac{32}{4} =8 $
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