$S$ , $P$ are sum and product of real roots of $x^4-7x^2-5=0$ what is the value of $2P^2-3SP+2S$? 
If $S$ and $P$ are sum and product of real roots of $x^4-7x^2-5=0$
what is the value of $2P^2-3SP+2S$ ?
$1)59-7\sqrt{69}\qquad\qquad2)7+\sqrt{69}\qquad\qquad3)50\qquad\qquad4)59+7\sqrt{69}$

I solved this problem as follow:
Since $f(x)=x^4-7x^2-5$ is an even function, sum of the roots is equal to zero Hence we are looking for the value of $2P^2$.
$$x^2=\dfrac{7\pm\sqrt{69}}{2}\quad\text{Since $x^2\ge0$, $\quad x^2=\dfrac{7+\sqrt{69}}{2}$ is acceptable}$$
$$x_1=+\sqrt{\dfrac{7+\sqrt{69}}{2}}\quad\text{,}\quad x_2=-\sqrt{\dfrac{7+\sqrt{69}}{2}}$$
$$2P^2=2\times \left(\dfrac{7+\sqrt{69}}{2}\right)^2=\dfrac{49+69+14\sqrt{69}}{2}=59+7\sqrt{69}$$
Is my answer right? Also I'm very eager to see if it is possible to solve this with other approaches.

Observe that $f$ is a function of the quadratic in $z = x^2$:  $$f(x) = z^2 - 7z - 5.$$  This has roots $$x^2 = z = \frac{7 \pm \sqrt{(-7)^2 - 4(-5)}}{2} = \frac{7 \pm \sqrt{69}}{2}.$$  Since the negative root for $z$ yields non-real roots for $x$, we can discard these, and find that the two real roots of $f$ are $$x \in \pm \sqrt{\frac{7 + \sqrt{69}}{2}}.$$  Their sum $S$ is indeed zero, and their product is simply $$P = - \frac{7 + \sqrt{69}}{2}.$$  Thus $$2P^2 = 2 \left(\frac{7 + \sqrt{69}}{2}\right)^2 = \frac{1}{2} (49 + 14\sqrt{69} + 69) = 59 + 7 \sqrt{69}.$$  This is nearly identical to your solution.
