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!
 A: You can use Kahan's method, see Help finding x in cubic function
For your example, let $f(x) = Ax^3+Bx^2+Cx+D = 2x^3-24x^2+23x+283$
$k = -{B\over3A} = {24\over6} = 4$ 
$f(k)/A = 119/2$ 
$f'(k)/A = -73/2 < 0$ 
$guess = k - (1.324718)(1)max(\sqrt[3]{119/2}, \sqrt{73/2}) = -4.00331$
Newton's method:
$-4.00331 → -2.96890 → -2.74377 → -2.73328 → -2.73326 → x = -2.73325591833$

$|x^3| ≈ 20.4193 < |D/A| = 283/2$ → deflate cubics using synthetic division
$E=Ax+B=-29.4665118367$
$F=Ex+C=103.53951787$ 
For quadratics, $Ax^2+Ex+F=0$, go for the "big" root, to reduce errors.  
$\large x_1 = -{E\;+\;sign(E) \sqrt{E^2-4AF} \over 2A} = \normalsize 8.94695974511$
$\large x_2 = {F/A\over x_1} = \normalsize 5.78629617322$ 
$$f(x)\;roots = -2.73325591833, 5.78629617322,8.94695974511$$
A: 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
$$x_1=4+\frac{7}{\sqrt{2}} \approx 8.94975 \quad x_2=4+\frac{7 \left(\sqrt{3}-1\right)}{2 \sqrt{2}}\approx 5.81173\quad x_3=4-\frac{7 \left(\sqrt{3}+1\right)}{2 \sqrt{2}}\approx -2.76148$$ 
A: Numerical methods result in three real solutions. 
Here are decimal approximations to the roots. 
$$-2.733256, 5.7862962,8.9469597$$
