Question 391155
I will assume that your function is {{{y = (3+x^2)/(4-x^2)}}}.

To find the range solve for x in terms of y and then state conditions on y that may arise.
{{{y = (3+x^2)/(4-x^2)}}}
<==> {{{4y - yx^2 = 3 + x^2}}}
<==> {{{0 = (1+y)x^2 - 4y + 3}}}
==> {{{x = (0 +- sqrt(-4*(1+y)*(3-4y) ))/(2*(1+y)) = ( 0+- sqrt( (4y - 3)(1+y)))/(1+y) }}}, using the quadratic formula.
Since the values of x must be real, the conditions we should impose on y are 

{{{ (1+y)*(3-4y)>= 0}}} and y is not equal to -1.
The solution set to this inequality is the union of intervals ({{{-infinity}}}, -1)U[3/4, {{{infinity}}}).  This is the RANGE of the function.