Answer
Solution: $\{1,2\}$
Work Step by Step
$(4-x)^{2}-5(4-x)+6=0$
Let $u= (4-x) $ we get,
$u^{2}-5u+6=0$
Splitting the middle term into two numbers whose product is $6$ and whose sum is, $-5$. The numbers are $-3$ and $-2$
$u^{2}-3u-2u+6=0$
$u(u-3)-2(u-3)=0$
$(u-3)(u-2)=0$
$u= 3$ or $u=2$
Substituting $u$ values in $u= (4-x) $ to find $x$
$u= 4-x $
$3=4-x$
$x = 4-3$
$x = 1$
$2=4-x$
$x = 4-2$
$x= 2$
Solution: $\{1,2\}$
