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3p^2=13
We move all terms to the left:
3p^2-(13)=0
a = 3; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·3·(-13)
Δ = 156
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{39}}{2*3}=\frac{0-2\sqrt{39}}{6} =-\frac{2\sqrt{39}}{6} =-\frac{\sqrt{39}}{3} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{39}}{2*3}=\frac{0+2\sqrt{39}}{6} =\frac{2\sqrt{39}}{6} =\frac{\sqrt{39}}{3} $
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