28=2p^2+5

Solution for 28=2p^2+5 equation:

28=2p^2+5

We move all terms to the left:

28-(2p^2+5)=0

We get rid of parentheses

-2p^2-5+28=0

We add all the numbers together, and all the variables

-2p^2+23=0

a = -2; b = 0; c = +23;
Δ = b2-4ac
Δ = 02-4·(-2)·23
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{46}}{2*-2}=\frac{0-2\sqrt{46}}{-4}
=-\frac{2\sqrt{46}}{-4}
=-\frac{\sqrt{46}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{46}}{2*-2}=\frac{0+2\sqrt{46}}{-4}
=\frac{2\sqrt{46}}{-4}
=\frac{\sqrt{46}}{-2}
$