Finding the equations of the tangents where a quadratic equation cuts the $x$-axis and the angle between the tangents (differentiation involved) Calculate the equations of the tangent where $y=x^2-5x-24$ cuts the $x$-axis. 

$(x-8)(x+3)$ factorising

$x=8,  x=-3 $

$y'(x)=2x-5$

$y'(8)=11$

$y'(-3)=-11$

$y=11x+c$

$0=11(8)+c$

And then I find, $c$, and repeat for the other tangent equation which gives:

$y=11x-88$ and $y=-11x-33$

The second question is what is the angle between the tangents and I don't know how to find it. I know it has something to do with tan-theta.
Could someone also check if my arithmetic is correct for the first part.
Notice, the slope of tangent at general point of the curve $y=x^2-5x-24$ $$\frac{dy}{dx}=\frac{d}{dx}(x^2-5x-24)=2x-5$$ Now, the point where curve  $y=x^2-5x-24$ cuts the x-axis has $y=0$ thus we have $$0=x^2-5x-24$$ Solving the quadratic equation for the values of $x$ as follows  $$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-24)}}{2(1)}$$ $$=\frac{5\pm 11}{2} \iff x=8, -3$$ Hence, we get two points $(8, 0)$ &amp; $(-3, 0)$ where curve intersects x-axis 

Now, the slope of the tangent at $(8, 0)$ is $$=\frac{dy}{dx}|_{x=8}=2\times 8-5=11$$ Hence its equation $$y-0=11(x-8)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=11x-88}}$$

Similarly, the slope of the tangent at $(-3, 0)$ is $$=\frac{dy}{dx}|_{x=-3}=2\times (-3)-5=-11$$ Hence its equation $$y-0=-11(x-(-3))$$ 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{y=-11x-33}}$$
Hence, the angle between the tangents is given as $$\tan \theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ now, setting $m_1=11$ &amp; $m_2=-11$, we get
$$\tan \theta=\left|\frac{11-(-11)}{1+11(-11)}\right|$$ 
$$=\left|\frac{11}{-60}\right|=\frac{11}{60}$$  Hence, $$\theta=\tan^{-1}\left(\frac{11}{60}\right)$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{acute angle:}\ \theta=\tan^{-1}\left(\frac{11}{60}\right)\approx10.39^\circ}}$$

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{obtuse angle:}\ \theta=\pi-\tan^{-1}\left(\frac{11}{60}\right)\approx 169.61^\circ}}$$
Basically the slope or the coeffecient of x in the tangent equation represents tan(theta)

So what you need to do is get artan(-11)= -84.81 then arctan(11)=84.81 and find the difference =169.62 deg you might want to take use the acute angel 10.39 deg