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3p^2=36
We move all terms to the left:
3p^2-(36)=0
a = 3; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·3·(-36)
Δ = 432
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*3}=\frac{0-12\sqrt{3}}{6} =-\frac{12\sqrt{3}}{6} =-2\sqrt{3} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*3}=\frac{0+12\sqrt{3}}{6} =\frac{12\sqrt{3}}{6} =2\sqrt{3} $
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