Example 1 :
The method can be used for the following equation :
$$\frac{abc-a^2}{c^2}x^{4n}+ax^{3n}+bdx^{2n}+cdx^n+d^2=0\tag1$$
where $n$ is a positive integer, and $a,b,c,d$ are real numbers with $c\not=0$.
Since $(1)$ can be written as
$$d^2+d(bx^{2n}+cx^n)+\frac{abc-a^2}{c^2}x^{4n}+ax^{3n}=0$$
which can be seen as a quadratic equation in $d$, we see that the discriminant is
$$\begin{align}&(bx^{2n}+cx^n)^2-4\bigg(\frac{abc-a^2}{c^2}x^{4n}+ax^{3n}\bigg)
\\\\&=\bigg(\frac{bc-2a}{c}\bigg)^2x^{4n}+2(bc-2a)x^{3n}+c^2x^{2n}
\\\\&=\bigg(\dfrac{bc-2a}{c}x^{2n}+cx^n\bigg)^2\end{align}$$
from which we obtain
$$d=\frac{-bx^{2n}-cx^n\pm\bigg(\dfrac{bc-2a}{c}x^{2n}+cx^n\bigg)}{2}$$
which are quadratic equations in $x^n$.
Example 2 :
This is an expansion of @Micah's comment.
The method can be used for the following equation :
$$\sqrt{ax^n+b}\ =-(a-c)^2x^{2n}+cx^n+b\tag2$$
where $n$ is a positive integer, and $a,b,c$ are real numbers.
(Micah's example $5-x^2=\sqrt{5-x}\ $ is the case where $(n,a,b,c)=(1,-1,5,0)$.)
Squaring the both sides of $(2)$ gives
$$b^2-\bigg(2(a-c)^2x^{2n}-2cx^n+1\bigg)b+(a-c)^4x^{4n}-2c(a-c)^2x^{3n}+c^2x^{2n}-ax^n=0$$
which can be seen as a quadratic equation in $b$.
The discriminant is
$$\begin{align}&\bigg(2(a-c)^2x^{2n}-2cx^n+1\bigg)^2-4\bigg((a-c)^4x^{4n}-2c(a-c)^2x^{3n}+c^2x^{2n}-ax^n\bigg)
\\\\&=4(a-c)^2x^{2n}+4(a-c)x^n+1
\\\\&=(2(a-c)x^n+1)^2\end{align}$$
so we get
$$b=\frac{2(a-c)^2x^{2n}-2cx^n+1\pm(2ax^n-2cx^n+1)}{2}$$
which are quadratic equations in $x^n$.
Example 3 :
The method can be used for the following equation :
$$\sqrt{ax^n+b}\ +\sqrt{\frac{(2da-a+c)^2}{2d}x^{2n}+cx^n+d^2-2bd+b}\ =d\tag3$$
where $n$ is a positive integer, and $a,b,c,d$ are real numbers with $d\gt 0$.
Squaring the both sides of
$$\sqrt{\frac{(2ad-a+c)^2}{2d}x^{2n}+cx^n+b+d^2-2bd}\ =d-\sqrt{ax^n+b}$$
gives
$$\frac{(2ad-a+c)^2}{2d}x^{2n}+cx^n+b+d^2-2bd=d^2-2d\sqrt{ax^n+b}\ +ax^n+b$$
which can be simplified as
$$\sqrt{ax^n+b}\ =-\bigg(a-\frac{a-c}{2d}\bigg)^2x^{2n}+\frac{a-c}{2d}x^n+b$$
This is of the form $(2)$, so treating this as a quadratic equation in $b$ works.
If we want to solve
$$4x^{20}-4x^{15}-15x^{10}+3x^5+9=0$$
which is of the form $(1)$, then we have
$$\begin{align}&3^2+3(-5x^{10}+x^5)+4x^{20}-4x^{15}=0
\\\\&\iff 3=\frac{5x^{10}-x^5\pm \sqrt{(3x^{10}+x^5)^2}}{2}
\\\\&\iff 3=4x^{10}\qquad \text{or}\qquad 3=x^{10}-x^5
\\\\&\iff x^{10}=\frac 34\qquad\text{or}\qquad x^5=\frac{1\pm\sqrt{13}}{2}
\\\\&\iff x=\pm\sqrt[10]{\frac 34},\ \sqrt[5]{\frac{1+\sqrt{13}}{2}},\ -\sqrt[5]{\frac{\sqrt{13}-1}{2}}\end{align}$$
If we want to solve
$$\sqrt{2x^4+7}\ =-x^{8}+x^4+7$$
which is of the form $(2)$, then squaring the both sides, we get
$$\begin{align}&7^2+7(-2x^8+2x^4-1)+x^{16}-2x^{12}+x^8-2x^4=0
\\\\&\implies 7=\frac{2x^8-2x^4+1\pm\sqrt{(2x^4+1)^2}}{2}
\\\\&\implies 7=x^8+1\qquad\text{or}\qquad 7=x^8-2x^4
\\\\&\implies x^4=\pm\sqrt 6,\ 1\pm 2\sqrt{2}\end{align}$$
It follows from $x^4\in\bigg[0,\dfrac{1+\sqrt{29}}{2}\bigg)$ that the solutions are $$x=\pm \sqrt[8]6$$
If we want to solve
$$\sqrt{20x^{6}+x^3-14}\ +\sqrt{x^3+6}\ =10$$
which is of the form $(3)$, then squaring the both sides of
$$\sqrt{20x^{6}+x^3-14}\ =10-\sqrt{x^3+6}$$
gives
$$\sqrt{x^3+6}\ =-x^6+6$$
which is of the form $(2)$. Squaring the both sides, we obtain
$$\begin{align}&6^2+6(-2x^6-1)+x^{12}-x^3=0
\\\\&\implies 6=\frac{2x^6+1\pm\sqrt{(2x^3+1)^2}}{2}
\\\\&\implies 6=x^6+x^3+1\qquad\text{or}\qquad 6=x^6-x^3
\\\\&\implies x^3=\frac{-1\pm\sqrt{21}}{2},\ \frac{1\pm 5}{2}\end{align}$$
It follows from $x^3\in \bigg[-\sqrt 6,\dfrac{-1-\sqrt{1121}}{40}\bigg]\cup\bigg[\dfrac{-1+\sqrt{1121}}{40},\sqrt 6\bigg]$ that the solutions are $$x=-\sqrt[3]2,\ \ \sqrt[3]{\frac{-1+\sqrt{21}}{2}}$$
