Solve p^4-12p^2-409b=0

$$p ^ { 4 } - 12 p ^ { 2 } - 409 b = 0$$

Solve for b

$$b=\frac{p^{2}\left(p^{2}-12\right)}{409}$$

Steps for Solving Linear Equation

Subtract p^{4} from both sides. Anything subtracted from zero gives its negation.

$$-12p^{2}-409b=-p^{4}$$

Add 12p^{2} to both sides.

$$-409b=-p^{4}+12p^{2}$$

The equation is in standard form.

$$-409b=12p^{2}-p^{4}$$

Divide both sides by -409.

$$\frac{-409b}{-409}=\frac{p^{2}\left(12-p^{2}\right)}{-409}$$

Dividing by -409 undoes the multiplication by -409.

$$b=\frac{p^{2}\left(12-p^{2}\right)}{-409}$$

Divide \left(-p^{2}+12\right)p^{2} by -409.

$$b=\frac{p^{4}-12p^{2}}{409}$$

Solve for p (complex solution)

$$p=\sqrt{-\sqrt{409b+36}+6}$$

$$p=-\sqrt{-\sqrt{409b+36}+6}$$

$$p=-\sqrt{\sqrt{409b+36}+6}$$

$$p=\sqrt{\sqrt{409b+36}+6}$$

Solve for p

$$\left\{\begin{matrix}p=-\sqrt{\sqrt{409b+36}+6}\text{; }p=\sqrt{\sqrt{409b+36}+6}\text{, }&b\geq -\frac{36}{409}\\p=-\sqrt{-\sqrt{409b+36}+6}\text{; }p=\sqrt{-\sqrt{409b+36}+6}\text{, }&b\geq -\frac{36}{409}\text{ and }b\leq 0\end{matrix}\right.$$