Determine all real roots of the equation x^2-9=-6/(x^2-4)?

First, we'll re-write the number 9, from the left side,
as - 4 - 5 We'll re-write the equation:

x^2 - 4 - 5 =
-6/(x^2-4)

We notice that we've created the structure x^2 -
4.

We'll note x^2 - 4 = t

t  -
5 = -6/t

We'll multiply by t both
sides:

t^2 - 5t  + 6 = 0

We'll
apply quadratic formula:

t1 = [5 + sqrt(25 -
24)]/2

t1= (5+1)/2

t1
=3

t2 = (5-1)/2

t2 =
2

We'll put x^2 - 4 = t1 => x^2 - 4 =
3

x^2 = 7

x1 = sqrt7 and x2 =
-sqrt7

We'll put x^2 - 4 = t2 => x^2 - 4=
2

x^2 =6

x3 = sqrt6 and x4 =
-sqrt6

The all 4 real solutions of the
equation are: {-sqrt7 ; -sqrt6 ; sqrt6 ; sqrt7}.