Solve the equation algebraically, and then use a calculator to find the values on the interval $[0,2 \pi$ ). (Enter your answers as a comma-separated list. Round your answers to four decimal places.) \[ \tan ^{2}(x)+8 \tan (x)-11=0 \]

Step 1 ：By the quadratic formula, we have \[\tan x = \frac{-8 \pm \sqrt{64+44}}{2} = \frac{-8 \pm \sqrt{108}}{2} = \frac{-8 \pm 2\sqrt{27}}{2} = -4 \pm \sqrt{27}.\] Let \(r_1 = -4 + \sqrt{27}\) and \(r_2 = -4 - \sqrt{27}\). Note that \(r_1 r_2 = -11\).

Step 2 ：Graphing \(y = \tan x\), we see that \(\tan x = r_1\) for two angles in \([0,2 \pi]\), and \(\tan x = r_2\) for two angles in \([0,2 \pi]\).

Step 3 ：Let \(\alpha = \arctan r_1\), and let \(\beta = \arctan r_2\), which are two of the solutions. Note that \[\tan \left( \frac{\pi}{2} - \alpha \right) = \frac{\sin (\frac{\pi}{2} - \alpha)}{\cos (\frac{\pi}{2} - \alpha)} = \frac{\cos \alpha}{\sin \alpha} = \frac{1}{\tan \alpha} = \frac{1}{r_1} = r_2.\] It follows that \(\beta = \frac{\pi}{2} - \alpha\), or \[\alpha + \beta = \frac{\pi}{2}.\]

Step 4 ：The other two solutions are \(\alpha + \pi\) and \(\beta + \pi\). Hence, the four solutions are \(\alpha\), \(\beta\), \(\alpha + \pi\), and \(\beta + \pi\).

Step 5 ：Using a calculator, we find that \(\alpha \approx 1.9823\), \(\beta \approx -0.1593\), \(\alpha + \pi \approx 5.1240\), and \(\beta + \pi \approx 2.9823\).

Step 6 ：However, since \(\beta\) is not in the interval \([0,2 \pi)\), we discard it. Thus, the solutions are \(\alpha\), \(\alpha + \pi\), and \(\beta + \pi\).

Step 7 ：Rounding to four decimal places, the solutions are \(\boxed{1.9823, 2.9823, 5.1240}\).