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3(p+6)p=12
We move all terms to the left:
3(p+6)p-(12)=0
We multiply parentheses
3p^2+18p-12=0
a = 3; b = 18; c = -12;
Δ = b2-4ac
Δ = 182-4·3·(-12)
Δ = 468
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{13}}{2*3}=\frac{-18-6\sqrt{13}}{6} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{13}}{2*3}=\frac{-18+6\sqrt{13}}{6} $
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