#### Answer

$4$

#### Work Step by Step

$x-\sqrt (17-4x) -3 = 0$
$x-3 = \sqrt (17-4x) $
Squaring on both sides.
$(x-3)^{2} = 17-4x$
Using $(a-b)^{2} = a^{2} -2ab +b^{2} $
$(x-3)^{2} = 17-4x$
$x^{2}-6x+9= 17-4x$
$x^{2}-6x+9- 17+4x=0$
$x^{2}-2x-8=0$
By factoring,
$(x-4)(x+2)=0$
$x=4$ or $x=-2$
Substituting $x=4$ and $x=-2$ in the given equation, $x=4$ only satisfies the equation. $x=-2$ does not satisfy. So, $x=4$ is the solution.