Pick some point $(a,b)$ which lies on the curve $y=\frac{x}{x+1}$. This means that $b=\frac{a}{a+1}$. Taking the derivative using the quotient rule, we find:
$$y'(x)=\frac{1}{(1+x)^2}$$
so the slope of the tangent line at our point $(a,b)$ is
$$y'(a)=\frac{1}{(1+a)^2}$$
The point slope formula then tells us that the equation of the tangent line to the curve at the point $(a,b)$ is
$$y-b=\frac{1}{(1+a)^2}(x-a)$$
We want the point $(1,2)$ to satisfy this equation, so remembering that $b=\frac{a}{a+1}$, we have:
$$2-b=2-\frac{a}{a+1}=\frac{1}{(1+a)^2}(1-a)$$
This equation reduces to the quadratic $a^2+4a+1=0$ (which is the quadratic you found). The solutions $a=-2\pm\sqrt{3}$ can be plugged back into the equation
$$y=\frac{1}{(1+a)^2}(x-a)+\frac{a}{1+a}=\frac{1}{(1+a)^2}x+\frac{a^2}{(1+a)^2}$$
to obtain the equations of the desired tangent lines.
Here is a helpful picture plotted with Wolfram|Alpha:
Notice where the tangent lines intersect!
To find the slope of the curve, or the gradient you can differentiate the function and plug in the $x$ value of the point into the derivative and that will yield the gradient at that point. So you have $y=5-x^2$ at the point $(1, 4)$.
$$y'=-2x$$
The gradient of the curve, $y=5-x^2$ at $(1,4)$ is:
$$y'(1)=-2(1)=-2$$
For the equation of the tangent line, you have the gradient $(-2)$, a point on the line $(1, 4)$, and you know that the equation of a line is $y=mx+c$. You can plug in the values that you know to find the value of $c$ and then you'll have the equation of the tangent line.
Best Answer
The derivative is correct.
As stated in the execise, the point $(1,2)$ is not on the curve; so you need to find such a point $(x_0,y_0)$ which a) lies on the curve; b) the tangent line to the curve in this point passes through $(1,2)$. Are these hints sufficient?
Edit
Our line is described by the equation $y=kx+c$. It is tangent to the curve int he point $(x_0,y_0)$, thus $y_0=kx_0+c$; in addition, the derivatives of the line and of the curve must coincide in this point, i.e. $k=\frac{1}{(x_0+1)^2}$. Finally, the line passes though $(1,2)$, hence $2= k+c$.
Thus, we obtain the equations: $$y_0=kx_0+c\\y_0=x_0/(x_0+1)\\k=\frac{1}{(x_0+1)^2}\\2= k+c$$
Can you solve these equations to obtain all possible pairs $(k,c)$?