polynomials within $R$ will be discussed
$\mathbb{{D}}:R$
a degree two polynomial can be defined as:
$\displaystyle\frac{\delta}{\delta{a,b,x}}y=
\left(\begin{eqnarray}
a \\
x^2 \\
\end{eqnarray}\right)
\oplus
\left(\begin{eqnarray}
b \\
x \\
\end{eqnarray}\right)
\oplus{c}$
where:
$\{a,x\mid{a,x}\neq0\}\land\{c\mid(c\in{\bar{\mathbb{Z}}})\lor{c}\in\bar{d}\}$
the same conditions that apply to the amount of solutions for a
linear function also apply to a quadratic, however there are now two $x$-terms, meaning that the amount of solutions are now $2(n!)$, so there should be four solutions
to have a consistent way to find the solutions, however, a quadratic formula that's compatible with $R$ must be found. however, because multiplication is not communtative and because the process of deriving the quadratic formula requires a far amount of multiplication, new notation shall be made to make it so there aren't an exorbitant amount of formulae. this notation is the following:
$[x]^{\dot{\Box}{n}_m}$ or $[x]$
which shows that, for a number $x$, it may be multiplied by another number $n$ $m$ different ways that would result in a different product and that any of these combinations may happen without regard for any other, but with a connection to any other term with the same symbol in place of $\dot{\Box}$. the more simplified notation with just square brackets is used when there is no added multiplicand, but the multipliers themselves may be multiplied in various different ways. a subscript may be added, but the combinations may generally be inferred. this is done to condense clutter as much as possible. this notation should not be mistaken for representing a matrix
knowing this, the quadratic formulae may be derived as follows:
$\displaystyle\frac{\delta}{\delta{a,b,x}}0=
\left(\begin{eqnarray}
a \\
x^2 \\
\end{eqnarray}\right)
\oplus
\left(\begin{eqnarray}
b \\
x \\
\end{eqnarray}\right)
\oplus{c}$
$\displaystyle\frac{\delta}{\delta{x}}0=x^2a\oplus{x}b\oplus{c}$
$\displaystyle\frac{\delta}{\delta{x}}0=[x^2a]^{a_1^{{-1}}}\oplus[xb]^{a_3^{{-1}}}\oplus[c]^{a_2^{{-1}}}$
$\displaystyle\frac{\delta}{\delta{x}}0=[x^2a]^{a_1^{{-1}}}\oplus([b2^{-1}]^{a_2^{-1}})^2\oplus[xb]^{a_3^{{-1}}}\oplus[c]^{a_2^{{-1}}}\ominus([b2^{-1}]^{a_2^{-1}})^2$
$\displaystyle\frac{\delta}{\delta{x}}0=[([x]^{a_1^{-1}}\oplus[b2^{-1}]^{a_2^{-1}})^2]^{x_1}\oplus(([4ac]^{a_1}4^{-1}a^{-2})\ominus(b^24^{-1}a^{-2}))$
$\displaystyle\frac{\delta}{\delta{x}}(b^2\ominus4^{-1}[ac])4^{-1}a^{-2}=[([x]^{a_1^{-1}}\oplus{b}2^{-1}a^{-2})^2]^{x_1}$
$\displaystyle\frac{\delta}{\delta{x}}\pm\sqrt{(b^2\ominus4^{-1}[ac])4^{-1}a^{-2}}=[[x]^{a_1^{-1}}\oplus{b}2^{-1}a^{-2}]^{x_1}$
$\displaystyle\frac{\delta}{\delta{x}}-[b2^{-1}a^{-2}]^{\dot{x}_1}\pm(\sqrt{(b^2\ominus4^{-1}[ac])})2^{-1}a^{-1}=[[x]^{a_1^{-1}}]^{\dot{x}_1$
$\displaystyle-([b2^{-1}]^{\dot{x}_1}\pm\sqrt{(b^2\ominus4^{-1}[ac]})2^{-1}a^{-1}=[[x]^{a_1^{-1}}]^{\dot{x}_1$
$\Box$
that is the quadratic formula that is compatible with ring $R$. there are three terms that have square bracket notation:
$[b2^{-1}]^{\dot{x}_1}\;,\;[ac]\;,\;[[x]^{a_1^{-1}}]^{\dot{x}_1}
there are two ways to multiply term $ac$, one other way to multiply term $b2^{-1}$, and two ways to multiply term $x$; this is because the combination with $x$ between terms $b2^{-1}$ and $x$ are shared, notated with $\dot{x}_1$. this confirms the idea that there are four total solutions to a quadratic within ring $R$ as there are four total combinations for the terms within the quadratic formula, supporting the conjecture that the amount of solutions to an equation within ring $R$ has some relation to the factorial of its terms' multipliers and multiplicands