#### Answer

The four real solutions are:
$x = -2,~~$ $x=2,~~$ $x=-3,~~$ or $~~x=3$

#### Work Step by Step

Let's consider a trinomial in this form: $~x^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$
Then the next step is to rewrite the trinomial as follows:
$~x^2+bx+c = (x+r)~(x+s)$
We can consider the given equation with $x^2$ in place of $x$:
$x^4-13x^2+36 = 0$
To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = -13~$ and $~r\times s = 36$. We can see that $(-4)+(-9) = -13~$ and $(-4)\times (-9) = 36$
We can factor the equation as follows:
$x^4-13x^2+36 = 0$
$(x^2-4)~(x^2-9) = 0$
The general form of factoring the difference of two squares is as follows:
$x^2-y^2 = (x+y)~(x-y)$
We can use difference of squares to continue factoring the equation as follows:
$x^4-13x^2+36 = 0$
$(x^2-4)~(x^2-9) = 0$
$(x+2)~(x-2)~(x+3)~(x-3) = 0$
$x+2 = 0,~~$ $x-2=0,~~$ $x+3=0,~~$ or $~~x-3=0$
$x = -2,~~$ $x=2,~~$ $x=-3,~~$ or $~~x=3$