10=2p^2

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Solution for 10=2p^2 equation:



10=2p^2
We move all terms to the left:
10-(2p^2)=0
a = -2; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-2)·10
Δ = 80
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:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-2}=\frac{0-4\sqrt{5}}{-4} =-\frac{4\sqrt{5}}{-4} =-\frac{\sqrt{5}}{-1} $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-2}=\frac{0+4\sqrt{5}}{-4} =\frac{4\sqrt{5}}{-4} =\frac{\sqrt{5}}{-1} $

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