<s>[INST] Solving the equation $2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$ 
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 $t$.
 [/INST] 
$2t^4-4t^3-t^2+3t+1=0$

Alt. hint: $\;\;\displaystyle
2t^4-4t^3\color{red}{+2t^2-2t^2}-t^2+3t+1 = \underbrace{2t^2(t-1)^2-3t(t-1)+1}_{\small{\displaystyle 2u^2-3u+1=(u-1)(2u-1)}}
$
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