What are the real solutions of the equation y^2+3=13/(y^2-9) ?

First, we'll re-write the number 3, fro the left side, as
12 - 9. We'll re-write the equation:

y^2 - 9 + 12 = 13/(y^2
- 9)

We've made this change, to create the structure y^2 -
9.

We'll note y^2 - 9 = t

t +
12 = 13/t

We'll multiply by t both
sides:

t^2 + 12t - 13 =
0

We'll apply quadratic
formula:

t1 = [-12 + sqrt(144 +
52)]/2

t1= (-12 +
sqrt196)/2

t1 = (-12+14)/2

t1
=1

t2 = (-12-14)/2

t2 =
-13

We'll put y^2 - 9 = t1 => y^2 - 9 =
1

y^2 = 10

y1 = sqrt10 and y2
= -sqrt10

We'll put y^2 - 9 = t2 => y^2 - 9 =
-13

y^2 = -13+9

y^2 =
-4

Since there is no real value for y, that
raised to square to give -4, we'll conclude that the real solutions of the equation are:
{-sqrt10 ; sqrt10}.