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.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ),   by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i   0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5