Solve the inequality $x^4-3x^2+5\ge0$

Denote:

$x^4-3x^2+5 = a - 7/2$

Then:

$x^4-3x^2+12 = a + 7/2$

Now you want to solve this one:

$\sqrt{a-7/2} + \sqrt{a + 7/2} = 7$

This is quite symmetrical and nice to work with.
Raise it to power $2$ and proceed, should be trivial from there.

At the end do a direct check to see if the values you found for $a$ give rise to valid roots for $x$. For example, if you get a solution for $a$ smaller than $7/2$, that obviously does not work i.e. does not give you any solutions for $x$.

Hint :

First solve for $\sqrt{z} + \sqrt{z+7} = 7 $

Next substitute $z=x^4-3x^2+5$ and solve for $x$.

---

Actually easy to see $\sqrt{9} + \sqrt{9+7} = 7 $.

$9$ is only solution as monotonic function has unique root (when it does).

So only need to solve for $x^4-3x^2+5=9$.