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4p^2+12p-72=0
a = 4; b = 12; c = -72;
Δ = b2-4ac
Δ = 122-4·4·(-72)
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$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36}{2*4}=\frac{-48}{8} =-6 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36}{2*4}=\frac{24}{8} =3 $
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