A (in a narrower sense) is the special case of the quartic equation (http://planetmath.org/QuarticFormula) containing no odd degree terms:

 $\displaystyle ax^{4}+bx^{2}+c=0$ (1)

Here, $a$, $b$, $c$ are known real or complex numbers and  $a\neq 0$.

For solving a biquadratic equation (1) one does not need the quartic formula (http://planetmath.org/QuarticFormula) since the equation may be thought a quadratic equation with respect to $x^{2}$, i.e.

 $a(x^{2})^{2}+bx^{2}+c=0,$

whence

 $x^{2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

(see quadratic formula or quadratic equation in $\mathbb{C}$ (http://planetmath.org/QuadraticEquationInMathbbC)).  Taking square roots of the values of $x^{2}$ (see taking square root algebraically), one obtains the four roots (http://planetmath.org/Equation) of (1).

 $\displaystyle x^{4}+x^{2}-20=0.$ (2)

We have

 $\displaystyle x^{2}=\frac{-1\pm\sqrt{1^{2}-4\cdot 1\cdot(-20)}}{2\cdot 1}=% \frac{-1\pm 9}{2},$ (3)

i.e.  $x^{2}=4$  or  $x^{2}=-5$.  The solution is

 $\displaystyle x=\pm 2\quad\lor\quad x=\pm i\sqrt{5}.$ (4)

Remark.  In one wants to form of rational numbers a polynomial equation with rational coefficients and most possibly low degree by using two square root operations, then one gets always a biquadratic equation.  A couple of examples:

1) $x=1+\sqrt{2}+\sqrt{3}$
$(x-1)^{2}=2+2\sqrt{6}+3$
$y^{2}-5=2\sqrt{6}$
$y^{4}-10y^{2}+1=0$  (one has substituted (http://planetmath.org/TchirnhausTransformations)  $x-1:=y$)

2) $x=\sqrt{\sqrt{2}-1}$
$x^{2}=\sqrt{2}-1$
$(x^{2}+1)^{2}=2$
$x^{4}+2x^{2}-1=0$