Solve (D^3+4D^2-6)y=0

$$( D ^ { 3 } + 4 D ^ { 2 } - 6 ) y = 0$$

Solve for y

$$\left\{\begin{matrix}\\y=0\text{, }&\text{unconditionally}\\y\in \mathrm{R}\text{, }&D=\frac{4\left(2\cos(\frac{\arccos(\frac{17}{64})+4\pi }{3})-1\right)}{3}\text{ or }D=\frac{4\left(2\cos(\frac{\arccos(\frac{17}{64})+2\pi }{3})-1\right)}{3}\text{ or }D=\frac{4\left(2\cos(\frac{\arccos(\frac{17}{64})}{3})-1\right)}{3}\end{matrix}\right.$$

Steps for Solving Linear Equation

Divide 0 by D^{3}+4D^{2}-6.

$$y=0$$

Solve for D

$$\left\{\begin{matrix}\\D=\frac{4\left(-\sqrt{3}\sin(\frac{\arccos(\frac{17}{64})}{3})-\cos(\frac{\arccos(\frac{17}{64})}{3})-1\right)}{3}\approx -3.514136929\text{; }D=\frac{4\left(2\cos(\frac{\arccos(\frac{17}{64})}{3})-1\right)}{3}\approx 1.086130198\text{; }D=\frac{4\left(\sqrt{3}\sin(\frac{\arccos(\frac{17}{64})}{3})-\cos(\frac{\arccos(\frac{17}{64})}{3})-1\right)}{3}\approx -1.571993268\text{, }&\text{unconditionally}\\D\in \mathrm{R}\text{, }&y=0\end{matrix}\right.$$