#### Answer

$x = \{ 1, \frac{1}{4}\}$

#### Work Step by Step

$\sqrt (5x-1) -\sqrt x + 2 = 3 $
To separate the radicals add $-2$ both sides.
$\sqrt (5x-1) -\sqrt x +2 - 2 = 3 - 2 $
$\sqrt (5x-1) -\sqrt x = 1 $
Squaring on both sides.
$(\sqrt (5x-1) -\sqrt x)^{2} = 1^{2} $
$(5x-1) + x - 2 \sqrt (5x-1)\sqrt x = 1$
$6x-1 - 2 \sqrt (5x-1)\sqrt x = 1$
$6x-1 -1 = 2 \sqrt (5x-1)\sqrt x $
$6x-2 = 2 \sqrt (5x-1)\sqrt x $
$2(3x-1) = 2 \sqrt (5x-1)\sqrt x $
Dividing both sides by $2$
$(3x-1) = \sqrt (5x-1)\sqrt x $
Squaring on both sides.
$(3x-1)^{2} = (\sqrt (5x-1)\sqrt x )^{2}$
$9x^{2}-6x+1 = x(5x-1)$
$9x^{2}-6x+1 = 5x^{2}-x$
$9x^{2}-6x+1 - 5x^{2}+x = 0$
$4x^{2}-5x+1 = 0$
By factoring,
$(4x-1)(x-1)=0$
$x = 1$ or $x= \frac{1}{4}$
$x = \{ 1, \frac{1}{4}\}$
Substituting $x$ values in the given equation, $x= 1$ satisfies the equation. So, $1$ is the only solution.