Combining and simplifying equations I've come across a derivation in a journal article where one of the steps is to substitute one expression into another, but I cannot simplify to their solution. The Expressions are:
Expression 1
$$
\frac{d\theta_{t}}{dt} = k_{a}q_{max}X\theta_{t}^2 -(k_{a}C_{0} + k_{a}q_{max}X+k_{d})\theta_{t} +k_{a}C_{0}
$$
Expression 2
$$
\theta_{e} = \frac{k_{a} (q_{max}X + C_{0})+k_{d} - \sqrt{\Delta}}{2k_{a}q_{max}X}
$$
In the article, it says that Expression 2 is substituted into Expression 1, to obtain Expression 3:
Expression 3
$$
\frac{d\theta_{t}}{dt} = k_{1}(\theta_{e}-\theta_{t})+k_{2}(\theta_{e}-\theta_{t})^2
$$
With the definitions
$$
k_{1}=\sqrt{\Delta}
$$
$$
k_{2} = k_{a}q_{max}X
$$
Unfortunately the authors omitted the steps in between, so to check I initially simplified Expression 1 and Expression 2 using the definitions:
$$
\frac{d\theta_{t}}{dt} = k_{2}\theta_{t}^2 -(k_{a}C_{0} + k_{2}+k_{d})\theta_{t} +k_{a}C_{0}
$$
$$
\theta_{e} = \frac{k_{2}+ k_{a}C_{0}+k_{d} - k_{1}}{2k_{2}}
$$
Rearranging the second expression I get:
$$
k_{a}C_{0} =2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$
Which I substituted into the first expression to get
$$
\frac{d\theta_{t}}{dt} = k_{2}\theta_{t}^2 -(2k_{2}\theta_{e}+k_{1})\theta_{t} +2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$
Multiplying out the brackets and completing the square, I get:
$$
\frac{d\theta_{t}}{dt} =k_{2}(\theta_{t}-\theta_{e})^2 - k_{2}\theta_{e}^2 -k_{1}\theta_{t}+2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$
Which is where I become stuck. Is there a mathematical trick other than completing the square that I am missing which will allow me to simplify this expression to obtain Expression 3?
The paper that I have obtained this derivation from is by Lui and Shen, titled "From Langmuir Kinetics to First and Second Order Rate Equations for Adsorption"
Edit
In response to a comment, Delta is defined as:
$$
\Delta=k_{a}^2(C_{0}-q_{max}X)^2+2k_{a}k_{d}(C_{0}+q_{max}X)+k_{d}^2
$$
 A: Note that expression for $\theta_e$ looks suspiciously like a formula for a root of quadratic equation.  Specifically, it's the smaller of the two roots of the quadratic in the expression for $d\theta_t/dt$.  And the aptly named $\Delta$ is the quadratic's discriminant.
That's all we need to know.  Suppose $x_1$ and $x_2$ where $x_1 \le x_2$ are the two roots of
$$
ax^2 + bx + c
$$
Then the quadratic can be rewritten as
$$a(x-x_1)(x-x_2)$$
But, since $x_2-x_1 = \sqrt{\Delta}/a$ where $\Delta$ is the quadratic's discriminant, we can substitute $x_2 = x_1 + \sqrt{\Delta}/a$ to obtain
$$a(x_1-x)^2 + \sqrt{\Delta}(x_1-x)$$
Which is the expression you were trying to derive
A: To reduce visual clutter, I'm making a few definitions:
$$\theta:=\theta_t \qquad \theta':=\frac{d\theta_t}{dt} \qquad
\phi:=\theta_e \qquad q := q_{max} \qquad a := k_a \qquad d := k_d \qquad p := k_aC_0$$
Together with the provided definitions
$$k_1 := \sqrt{\Delta} \qquad k_2:=aqX$$
the equations become
$$\begin{align}
\theta' &= k_2\theta^2-(p+d+k_2)\theta+p \tag1\\[4pt]
2k_2\phi &= p+d-k_1+k_2 \tag2 \\[4pt]
\theta' &= k_1(\phi-\theta)+k_2(\phi-\theta)^2 \tag3 
\end{align}$$
As OP has encountered, simple algebraic manipulation won't get from $(1)$ and $(2)$ to $(3)$. It's required to know that
$$\begin{align}
\Delta &:= a^2(C_0-qX)^2+2ad(C_0+qX)+d^2 \tag{4} \\[4pt]
\quad\to\quad k_1^2 &=(p-k_2)^2+2d(p+k_2)+d^2 \tag{5}
\end{align}$$
There's probably a clever way to incorporate this relation, but a brute force way is to treat $(1)$ and $(2)$ as linear equations in $p$ and $d$, solve for these values, substitute into $(5)$, and simplify (dividing-through by $4k_2$). When the dust settles, we arrive at the target relation $(3)$.

Note. It's partially clever to rewrite $(5)$ as
$$\begin{align}
k_1^2 &= -4k_2p+(p+k_2)^2+2d(p+k_2)+d^2 \tag{6}\\[4pt]
\quad\to\quad 4k_2p &=(p+d+k_2)^2-k_1^2 \tag{7} \\[4pt]
&=(p+d-k_1+k_2)(p+d+k_1+k_2) \tag{8} \\[4pt]
\quad\to\quad 2p&=\phi(p+d+k_1+k_2) \tag{9}
\end{align}$$
(where we invoke $(2)$ to simplify the right-hand side of $(8)$). If we further define
$$m:=p+d+k_2 \tag{10}$$
then we have
$$\begin{align}
\theta' &= k_2\theta^2-m\theta+p\tag{1'} \\[4pt]
2k_2\phi &= m-k_1 \tag{2'} \\[4pt]
2p &= \phi(m+k_1) \tag{9'}
\end{align}$$
Eliminating $m$ and $p$ from the system to arrive at $(3)$ is then relatively straightforward.
