You are in school to learn.

My pal needs it, I'm a programmer with a job mate.

And I am the Queen of England.

Presuming you are in school, you need to know the process, not just the answer.

You have x^4 - 10x^2 + 9 = 0

There are two ways to solve this problem: 1) Factoring 2) Quadratic Equation + Substitution

FACTORING:
We will presume we can factor to:
(x^2 + a)(x^2 + b) = 0

We want to find 'a' and 'b' such that a*b = 9
and such that a + b = -10

Presuming that a and b are both integers, we have the following possibilities to satisfy a*b = 9
1, 9 : a + b = 10
-1,-9 : a + b = -10
3,3 : a + b = 6
-3,-3 : a + b = -6

You will notice that for a = -1 and b = -9 (or vice versa, order doesn't matter), a + b = -10 is satisfied.

We therefore have (x^2 - 1)(x^2 - 9) = 0
The roots for this equation are: x^2 = 1 and x^2 = 3

We take the square root of both sides, and we get:
x = -¦1 and x = -¦3
Note that we have 4 possible solutions in our solution set. This is because our polynomial is of degree 4. The Fundamental Theorem of Algebra allows us to conclude that every polynomial of degree n has exactly n (possibly complex) roots.

SUBSTITUTION + QUADRATIC FORMULA:
We have x^4 - 10x^2 + 9 = 0
Substitute t = x^2
t^2 - 10t + 9 = 0

Use:
a = 1, b = -10, c = 9
(+10 -¦ ((-10)^2 - 4*1*9)^(1/2)) / (2*1)

(10 -¦ (100 - 36)^(1/2)) / 2
(10 -¦ 8) / 2
18 / 2 or 2 / 2
t = 9 or t = 1

Since we substituted t = x^2, we have:
x^2 = 9 or x^2 = 1

The rest is the same as the factoring method above.

NOTE:
The factoring method that I presented made some critical assumptions, which are not always true: 1) that the roots of the polynomial were all integers 2) That you could factor the polynomial into a product of two binomials. Things get much more complicated if these two assumptions are not true. It all depends upon how cruel your textbook is.

Love 'adults' who are about as bright as 12 year olds.

Factor a trinomial.
(1 + -1x2)(9 + -1x2) = 0