(P^2-1)^2-11(p^2-1)+24=0

Solution for (P^2-1)^2-11(p^2-1)+24=0 equation:

(^2-1)^2-11(P^2-1)+24=0

We add all the numbers together, and all the variables

-11(P^2-1)=0

We multiply parentheses

-11P^2+11=0

a = -11; b = 0; c = +11;
Δ = b2-4ac
Δ = 02-4·(-11)·11
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-22}{2*-11}=\frac{-22}{-22}
=1
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+22}{2*-11}=\frac{22}{-22}
=-1
$