Solve for real y y^2+3+ 12/(y^2+3)=7

We'll note y^2 + 3 = t

We'll
re-write the equation in t:

t + 12/t =
7

We'll multiply by t both sides, after that, bringing all
terms to one side:

t^2 + 12 - 7t =
0

We'll re-arrange the
terms:

t^2 - 7t + 12 = 0

We'll
apply quadratic formula:

t1 = [7 + sqrt(49 -
48)]/2

t1 = (7+1)/2

t1 =
4

t2 = (7-1)/2

t2 =
3

But y^2 + 3 = t

For t =
4=> y^2 + 3 = 4

y^2 = 4 -
3

y^2 = 1

y1 = 1 and y2 =
-1

For t = 3 => y^2 + 3 =
3

y^2 = 0

y3 = y4 =
0

The real roots of the equation are: {-1 ; 0
; 1}.