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3p^2-18p=81=0
We move all terms to the left:
3p^2-18p-(81)=0
a = 3; b = -18; c = -81;
Δ = b2-4ac
Δ = -182-4·3·(-81)
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$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-36}{2*3}=\frac{-18}{6} =-3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+36}{2*3}=\frac{54}{6} =9 $
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