I have this equation
x^{(x^2-3x)}=x^2
I've found all roots except x = -1.
Can somebody help me?
Oops, I don't know how I could have gotten my factorisation wrong, sorry...
$\displaystyle \displaystyle \begin{align*} x^2 - 3x - 2 &= x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - 2 \\ &= \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{8}{4} \\ &= \left(x - \frac{3}{2}\right)^2 - \frac{17}{4} \\ &= \left(x - \frac{3}{2}\right)^2 - \left(\frac{\sqrt{17}}{2}\right)^2 \\ &= \left(x - \frac{3}{2} - \frac{\sqrt{17}}{2}\right)\left(x - \frac{3}{2} + \frac{\sqrt{17}}{2}\right) \end{align*}$