P+2=p^2-40

Solution for P+2=p^2-40 equation:

+2=P^2-40

We move all terms to the left:

+2-(P^2-40)=0

We get rid of parentheses

-P^2+40+2=0

We add all the numbers together, and all the variables

-1P^2+42=0

a = -1; b = 0; c = +42;
Δ = b2-4ac
Δ = 02-4·(-1)·42
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*-1}=\frac{0-2\sqrt{42}}{-2}
=-\frac{2\sqrt{42}}{-2}
=-\frac{\sqrt{42}}{-1}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*-1}=\frac{0+2\sqrt{42}}{-2}
=\frac{2\sqrt{42}}{-2}
=\frac{\sqrt{42}}{-1}
$