Solve the equation $$2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$$
Here is my work:
$$2x^2-5x+4=(4x-7)\sqrt{x-1}$$ $$2(x-1)^2-(x-1)+1=(4(x-1)-3)\sqrt{x-1}$$ Using the substitution $\sqrt{x-1}=t$, $$2t^4-t^2+1=t(4t^2-3)$$ $$2t^4-4t^3-t^2+3t+1=0$$ Here I plugged in some natural numbers like $1,2,3$ but neither of them worked. So I don't know how to factor the polynomial to find the values of $t$.
On
What you did is fine. Now, let\begin{align}p(t)&=2t^4-4t^3-t^2+3t+1\\&=2\left(t^4-2t^3-\frac12t^2+\frac32t+\frac12\right)\end{align}and let $r(t)$ be the depressed quartic of $t^4-2t^3-\frac12t^2+\frac32t+\frac12$, that is\begin{align}\left(t+\frac12\right)^4-2\left(t+\frac12\right)^3-\frac12\left(t+\frac12\right)^2+\frac32\left(t+\frac12\right)+\frac12&=t^4-2t^2+\frac{15}{16}\\&=\left(t^2-\frac54\right)\left(t^2-\frac34\right),\end{align}since the roots of the quadratic equation $T^2-2T+\frac{15}{16}=0$ are $\frac34$ and $\frac54$. So,\begin{align}p(t)&=2r\left(t-\frac12\right)\\&=2(t^2-t-1)\left(t^2-t-\frac12\right).\end{align}
Using substitution to get rid of a square root is a great approach. It's almost always better than squaring the equation. Well done!
An additional trick is if the resulting equation looks unpleasant, don't replace all the $x$ with $t^2+1$. Just replace the square root and leave the rest of the $x$s be.
Since it's not likely that anybody will ever ask you to solve a quartic equation the hard way, look for ways to then combine your equation with $t^2 = x - 1$ to create factoring opportunities.
$$2x^2 - 4xt + 7t - 5x + 4 = 0$$
Complete the square $2x^2 - 4xt$ by adding $2t^2 - 2x + 2=0$ and the equation becomes
$$2x^2 - 4xt + 2t^2 + 7t - 7x + 6=0$$
or
$$2(t-x)^2 + 7(t-x) + 6 = 0$$
This is now a quadratic for $t-x$ and the rest is simple.