Need help find the the real roots to this question

sqrt(5x-4)-sqrt(x)=2

$\begin{array}{|rcll|} \hline \sqrt{5x-4}-\sqrt{x} &=& 2 \quad &| \quad +\sqrt{x} \\ \sqrt{5x-4} &=& 2 +\sqrt{x} \quad &| \quad \text{square both sides} \\ 5x-4 &=& (2 +\sqrt{x})^2 \\ 5x-4 &=& 4+4\cdot \sqrt{x} + x \quad &| \quad -x \\ 4x-4 &=& 4+4\cdot \sqrt{x} \quad &| \quad :4 \\ x-1 &=& 1+ \sqrt{x} \quad &| \quad -1\\ x-2 &=& \sqrt{x} \quad &| \quad \text{square both sides} \\ (x-2)^2 &=& x\\ x^2-4x+4 &=& x \quad &| \quad -x \\ x^2-5x+4 &=& 0\\\\ x &=& \frac{5\pm \sqrt{25-4\cdot 4} } {2} \\ x &=& \frac{5\pm \sqrt{25-16} } {2} \\ x &=& \frac{5\pm \sqrt{9} } {2} \\ x &=& \frac{5\pm 3 } {2} \\\\ x_1 &=& \frac{5 + 3 } {2} \\ x_1 &=& \frac82 \\ \mathbf{x_1} & \mathbf{=} & \mathbf{4} \\\\ x_2 &=& \frac{5 - 3 } {2} \\ x_2 &=& \frac22 \\ \mathbf{x_2} & \mathbf{=} & \mathbf{1} \quad \text{ no solution}\\ \hline \end{array}$

$\sqrt{5x-4}-\sqrt x = 2\\ 5x-4+x-2(\sqrt{5x^2-4x})=4\\ 6x - 2\sqrt{5x^2-4x}=8\\ 3x-4=\sqrt{5x^2-4x}\\ 5x^2-4x=9x^2+16-24x\\ 4x^2-20x+16=0\\ x^2 - 5x + 4 = 0\\ (x-1)(x-4)=0\\ \color{blue}x=1\text{(rej.) OR }x=4\\ \color{red}x=4$