I need help with solving one Omar Al-Khayyam's distinct cubic equation, $x^3+ax^2=c$.

Question

The cubic equation of interest is $x^3+2x^2=6$. I found the two conics. The parabola is $y= x^2$ and the rectangular parabola, $(x+2)y=6$. I am familiar with the geometry of the parabola and can graph the rectangular hyperbola with the center at $(-2,0)$. I have been studying the works of Apollonius. In particular, the use of equal, area rectangles. I am graphing it geometrically and trying to see the relationship.

Answer 1

As Dietrich Burde commented, there is a formula for solving cubics. See here and here. A suggestion is rearranging to $x^3+2x^2-6=0$. Using Decartes' rule of signs, we also know there is 1 positive root which is $x \approx 1.34$. The other roots are complex as shown in the image below. Hope this helps, try using the formula. If you have trouble with that then I suggest using the Newton-Raphson method or substitution (harder)

Answer 2

Complex roots can be found by trigonometric function trick... $$x^3+2x^2-6=0.$$ Let $x=\frac{3}{2\sin t}$. Then, $$\frac{9}{4}+3\sin t-4\sin^3t=0$$ $$\frac94+\sin3t=0$$ $$t=-\frac13\arcsin(\tfrac{9}{4})$$ $$t=-\frac\pi 6+\frac i3\ln(\tfrac{\sqrt{65}+9}{4})$$ $$\sin t=...$$ So on. By hand it is complicated.

Edit: For the real root, we take $t=-\frac \pi 6+ \frac i3\ln(\tfrac{\sqrt{65}+9}{4})\color{red}{+\frac{2\pi}{3}}$ so the real root is $$x=\frac3{2\cosh\left(\frac1 3\ln(\tfrac{\sqrt{65}+9}{4})\right)}\approx 1.3402508$$