We know that the line is tangent at the point (2,8)
The derivative of the function at x=2 is
[tex]f'(x) = 6-2x \implies f'(2) = 6-4=2[/tex]
So, the tangent line passes through the point (2,8) and has slope 2. The equation is
[tex]y-8 = 2(x-2) \iff y = 2x+4[/tex]
This line crosses the x axis where y=0:
[tex]0 = 2x+4 \iff 2x = -4 \iff x = -2[/tex]
