Find C so that g(x) is continuous for all real numbers (10 Points!)

Hello i need help with this! i have an example in my notes but im not sure what the teacher was doing! please explain everything! 10points! Thanks in advance hope you understand how i put this togather lol.

Find c so that g(x)= { (2x^2+-x-15) / (x^2-8x+15), x<3 }
{cx+5 , x > or equal to 3 } is continuous for all real numbers.

A function is continuous at a if the limit as x approaches a from the right equals the limit as x approaches a from the left equals g(a), so we need to compute those 3 things.

g(3) = 3c+5
lim_{x-> 3+} g(x) = lim_{x->3+} cx+ 5 = 3c+5, since when x approaches from the right, it's always greater than 3.
lim_{x-> 3-} g(x) = lim_{x-> 3-} (2x^2+-x-15) / (x^2-8x+15) = lim_{x-> 3-} (2x +5)(x-3) / (x -5)(x-3) =
lim_{x-> 3-} (2x+5)/(x-5) = 11/-2 = -11/2

So we need -11/2 = 3c+5. Just solve for c:
-21/2 = 3c
-21/6 = c

g(x) = (x - 3)(2x + 5)/[(x - 3)(x - 5)] = (2x + 5)/(x - 5), x ≠ 3

g(3) = 11/(-2) = - 11/2

g(3) = 3c + 5

3c + 5 = -11/2
3c = -21/2
c = -7/2