y=(3x^2+2x-8)/(x^2+3) It should be clear that the denominator will never equal zero so we do not even need to consider it.
The numerator however can equal zero and those roots are what will be roots of the overall equation...
3x^2+2x-8=0
That is a quadratic of the form ax^2+bx+c. To factor it, you must first find two values that satisfy two conditions simultaneously.
ab=jk and b=j+k, in this case: jk=-24 and j+k=2 So the only values for j and k possible are -6 and 4
Now you replace the single linear term, bx, with jx and kx in the original equation...
3x^2-6x+4x-8=0 Now factor the 1st and 2nd pairs of terms...
3x(x-2)+4(x-2) which is equal to
(3x+4)(x-2)=0 so the values of x that make this true are:
x=-4/3 and x=2