This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

where
is a compact way to indicate both solutions
and
, while
are the coefficients of the quadratic equation, i.e. we consider the polynomial
.
So, in your case, we have 
Plug those values into the formula to get

So, the two solutions are


For
y = (x^2 -4)/((x +2)(x^2 -49))
the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving
y = (x -2)/(x^2 -49)
There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.
The horizontal asymptote is y=0.
The answer is b (I think not really sure)
Answer:Where is the picture?
Step-by-step explanation: