You can find it in Ramanujan's Notebooks IV by B. Berndt, Chap. 22 Elementary Results, Entry 20, p.31.
Ramanujan starts with this problem. Let $a,b,c,d$ be arbitrary. Solve the system,
$$x^2+ay = b\tag{20a}$$
$$y^2+cx = d\tag{20b}$$
Eliminating $y$, we find it is equivalent to,
$$a^2(d-cx) = (b-x^2)^2\tag{20.1}$$
Assume without loss of generality that $a=2$. Expanding this out,
$$x^4-2bx^2+4cx+(b^2-4d)=0\tag{20.1a}$$
By a simple linear substitution, the general quartic equation can be expressed in the depressed form,
$$x^4+px^2+qx+r = 0\tag{20.1b}$$
Equating coefficients of ${20.1a}$ and ${20.1b}$, you have a system of 3 equations in 3 unknowns {$b,c,d$}. Hence every quartic can be expressed in the form ${20.1}$. The problem then is to find $x$. Ramanujan defines,
$$x = \alpha+\beta+\gamma\tag{20.1c}$$
$$y = -(\alpha\beta+\beta\gamma+\gamma\alpha)$$
$$-c/2 = \alpha\beta\gamma$$
(If you are familiar with cubic equations, you'll already see where Ramanujan is going.)
Substitute $x,y,c$ into $(20a)$ and $(20b)$ keeping in mind $a=2$, then,
$$x^2+ay = \alpha^2+\beta^2+\gamma^2= b$$
$$y^2+cx = (\alpha\beta)^2+(\beta\gamma)^2+(\gamma\alpha)^2 = d$$
$$(-c/2)^2 = (\alpha\beta\gamma)^2$$
By elementary symmetric polynomials, we then conclude that $\alpha^2,\,\beta^2,\,\gamma^2$ are roots of the cubic equation,
$$t^3-bt^2+dt-c^2/4=0\tag{20.1d}$$
Of course, by solving $(20.1d)$, one can then find $\alpha,\,\beta,\,\gamma$. Using $(20.1c)$, we then further conclude that the four roots of the quartic are,
$$x = \alpha+\beta+\gamma, \quad\alpha-\beta-\gamma, \quad-\alpha-\beta+\gamma, \quad-\alpha+\beta-\gamma$$
P.S. This is similar to Euler's method where he solves a quartic as $x_i = \sqrt{y_1}\pm \sqrt{y_2}\pm \sqrt{y_3}$ and the $y_i$ are roots of a cubic. There's actually a generalization to this for solvable 8th deg eqns as $x_i = \sqrt{y_1}\pm \sqrt{y_2}\pm\dots\pm \sqrt{y_7}$ and the $y_i$ are roots of a 7th deg eqn. See this mathoverflow post.