I'll prove the following result:

$$K[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>\simeq K[s^3,s^2t,st^2,t^3],$$ where $K$ is a field.

Let $\varphi: K[x_1,x_2,x_3,x_4]\to K[s, t]$ be the ring homomorphism that maps $x_1\mapsto s^3$, $x_2\mapsto s^2t$, $x_3\mapsto st^2$ and $x_4\mapsto t^3$. Obviously $\operatorname{Im}\varphi=K[s^3,s^2t,st^2,t^3]$; this is a subring of $K[s,t]$ and the extension $K[s^3,s^2t,st^2,t^3]\subset K[s,t]$ is integral, hence $\dim K[s^3,s^2t,st^2,t^3]= \dim K[s,t]=2.$

It remains to prove that $\ker\varphi=\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>$. By definition $\varphi(f(x_1,x_2,x_3,x_4))=f(s^3,s^2t,st^2,t^3)$. In particular, this shows that the polynomials $g_1=x_1x_3-x_2^2$, $g_2=x_2 x_4-x_3^2$ and $g_3=x_1x_4-x_2 x_3$ belong to $\ker\varphi$.

Conversely, let $f\in\ker\varphi$, i.e. $f(s^3,s^2t,st^2,t^3)=0$. We want to show that $$f\in\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>.$$ The initial monomials of $g_1$, $g_2$, resp. $g_3$ with respect to the lexicographical order are $x_1x_3$, $x_2x_4$, resp. $x_1x_4$. The remainder on division of $f$ to $G=\{g_1,g_2,g_3\}$, denoted by $r$, is a $K$-linear combination of monomials none of which is divisible by $x_1x_3$, $x_2x_4$, resp. $x_1x_4$. This shows that the monomials of $r$ can have one the following forms: $x_1^ix_2^j$ with $i\ge 1$ and $j\ge 0$, $x_2^kx_3^l$ with $k\ge 1$ and $l\ge 0$, respectively $x_3^ux_4^v$ with $u,v\ge 0$. In order to prove that $f\in\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>$ it is enough to show that $r=0$. But we know that $f(s^3,s^2t,st^2,t^3)=0$ and therefore $r(s^3,s^2t,st^2,t^3)=0$. The monomials (in $s$ and $t$) of $r(s^3,s^2t,st^2,t^3)$ are of the following types: $s^{3i+2j}t^j$, $s^{2k+l}t^{k+2l}$, respectively $s^ut^{2u+3v}$. Now check that there is no possible cancelation between these monomials (because they can't be equal), so $r=0$.

Now it follows that $\dim R=2$.