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3p^2-12p-96=0
a = 3; b = -12; c = -96;
Δ = b2-4ac
Δ = -122-4·3·(-96)
Δ = 1296
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{1296}=36$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-36}{2*3}=\frac{-24}{6} =-4 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+36}{2*3}=\frac{48}{6} =8 $
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