Solving $2x^3-24x^2+23x+283=0$ I need to solve this equation:
$$\frac{x^2+1}{x-4}+\frac{x^2-1}{x+3}=23$$
When opened up it comes to this:
$$2x^3-24x^2+23x+283=0$$
Then I am stuck...
It feels like it has something to do with complex numbers perhaps? 
Any ideas? Thank you!

Let me be more puristic.
The first thing to notice is that the discriminant is positive $(\Delta=1582748)$; so three real roots with $p=-\frac{73}{2}$ and $q=\frac{119}{2}$.
Now, instead of Cardano, let us use the trigonometric method for cubic equation which write
$$x_k=-\frac b{3a}+2\sqrt{-\frac p 3} \cos\Big(\frac 13 \cos^{-1}\Big(\frac{3q}{2p}\sqrt{-\frac 3p } \Big)-\frac{2\pi}3k\Big)\qquad \text{for}\qquad k=0,1,2$$
Applied to your case this will give
$$x_k=4+\sqrt{\frac{146}{3}} \cos \left(\frac{2 \pi  k}{3}-\frac{1}{3} \cos
   ^{-1}\left(-\frac{357}{73}\sqrt{\frac{3}{146}}\right)\right)$$
Their decimal representation would be
$$x_0=8.94695974511457058 \qquad x_1=5.78629617321812462\qquad x_2=-2.73325591833269521$$
Edit
If you use your pocket calculator
$$\frac{1}{3} \cos
   ^{-1}\left(-\frac{357}{73}\sqrt{\frac{3}{146}}\right)\approx 0.782540 \qquad \text{while} \qquad \frac \pi 4\approx 0.785398$$ and $$\sqrt{\frac{146}{3}}\approx 6.97615$$ So, as an approximation,
$$x_k \sim 4+7 \cos \left(\frac{8 k-3}{12} \pi  \right)$$ which would give
$$x_1=4+\frac{7}{\sqrt{2}} \approx 8.94975 \quad x_2=4+\frac{7 \left(\sqrt{3}-1\right)}{2 \sqrt{2}}\approx 5.81173\quad x_3=4-\frac{7 \left(\sqrt{3}+1\right)}{2 \sqrt{2}}\approx -2.76148$$ 
