Answer
$7$
Work Step by Step
$\sqrt (x-3) + \sqrt (x+2) = 5$
Squaring on both sides to eliminate radicals.
$(\sqrt (x-3) + \sqrt (x+2))^{2} = 5^{2}$
$(x-3)+(x+2)+2\sqrt (x-3) \sqrt (x+2) = 25$
$2x-1+2\sqrt (x-3) \sqrt (x+2) = 25$
$2\sqrt (x-3) \sqrt (x+2) = 25+1-2x$
$2\sqrt (x-3) \sqrt (x+2) = 26-2x$
$2\sqrt (x-3) \sqrt (x+2) = 2(13-x)$
$\sqrt (x-3) \sqrt (x+2) = (13-x)$
Squaring on both sides to eliminate radicals.
$(\sqrt (x-3) \sqrt (x+2))^{2} = (13-x)^{2}$
$(x-3) (x+2) =169-26x+x^{2}$
$x^{2}-x-6 = 169-26x+x^{2}$
$x^{2}-x-6 - 169+26x-x^{2}=0$
$25x-175=0$
$25x=175$
$x = 7$
