Solve left(p^6+5p^3+12right)=0

$$\left( { p }^{ 6 } +5 { p }^{ 3 } +12 \right) =0$$

Solve for p

$$p\in \sqrt[3]{2}\sqrt[6]{3}e^{\frac{\left(\arctan(\frac{\sqrt{23}}{5})+\pi \right)i}{3}},\sqrt[3]{2}\sqrt[6]{3}e^{\frac{\arctan(\frac{\sqrt{23}}{5})i+5\pi i}{3}},\sqrt[3]{2}\sqrt[6]{3}e^{\frac{\arctan(\frac{\sqrt{23}}{5})i+3\pi i}{3}},\sqrt[3]{2}\sqrt[6]{3}e^{\frac{-\arctan(\frac{\sqrt{23}}{5})i+5\pi i}{3}},\sqrt[3]{2}\sqrt[6]{3}e^{\frac{-\arctan(\frac{\sqrt{23}}{5})i+\pi i}{3}},\sqrt[3]{2}\sqrt[6]{3}e^{\frac{-\arctan(\frac{\sqrt{23}}{5})i+3\pi i}{3}}$$

Steps Using the Quadratic Formula

Substitute t for p^{3}.

$$t^{2}+5t+12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 5 for b, and 12 for c in the quadratic formula.

$$t=\frac{-5±\sqrt{5^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$t=\frac{-5±\sqrt{-23}}{2}$$

Solve the equation t=\frac{-5±\sqrt{-23}}{2} when ± is plus and when ± is minus.

$$t=\frac{-5+\sqrt{23}i}{2} t=\frac{-\sqrt{23}i-5}{2}$$

Since p=t^{3}, the solutions are obtained by solving the equation for each t.

$$p=\sqrt[3]{2}\sqrt[6]{3}e^{\frac{-\arctan(\frac{\sqrt{23}}{5})i+5\pi i}{3}} p=\sqrt[3]{2}\sqrt[6]{3}e^{\frac{-\arctan(\frac{\sqrt{23}}{5})i+3\pi i}{3}} p=\sqrt[3]{2}\sqrt[6]{3}e^{\frac{-\arctan(\frac{\sqrt{23}}{5})i+\pi i}{3}} p=\sqrt[3]{2}\sqrt[6]{3}e^{\frac{\arctan(\frac{\sqrt{23}}{5})i+5\pi i}{3}} p=\sqrt[3]{2}\sqrt[6]{3}e^{\frac{\arctan(\frac{\sqrt{23}}{5})i+3\pi i}{3}} p=\sqrt[3]{2}\sqrt[6]{3}e^{\frac{\left(\arctan(\frac{\sqrt{23}}{5})+\pi \right)i}{3}}$$