F(x)=x^2+5x-5 points (0,f(0)) equation of the tangent line

Your post, "F(x)=x^2+5x-5 points (0,f(0)) equation of the tangent line," is a challenge to read.  Did you mean the following?

F(x)=x^2+5x-5.  Find the equation of the tangent line at (0, f(0))

If so, differentiate F(x) to obtain an expression for the slope of the tangent line:

F '(x) = 2x + 5

Now let x=0, to find the slope of the tangent line at (0,f(0)):

F '(0) = 2(0) + 5 = 5

Now you have the slope of the tangent line to the curve at (0, [0^2+5(0) - 5]).

Find the equation of the line with this slope that passes thru (0, -5):

y - [-5] = 5(x-[-5]),    or   y+5 = 5(x+5),    or y = -5 + 5x + 25.

Final  y = 5x + 20.

You just need to add 32 to 14 and get 46.

Um what? rephrase that ?