To solve the inequality
$x^2-4x+3 \gt 0$
, first factorise to give
$(x-1)(x-3) \gt 0$
.
Now sketch the graph of
$y=x^2-4x+3$
.

Because we are solving
$x^2-4x+3 \gt 0$
we want those values of
$y \gt 0$
.
From the graph we see this is true for
$x \lt 0$
or
$x \gt 3$
.
For the inequality
$x^2-4x+3 \lt 0$
, we need that part of the graph below the
$x$
axis. The solution is
$1 \lt x \lt 3$
.
To solve the inequality
$x^2-4x+3 \ge 0$
, just replace each 'greater than' sign with 'greater than or equal to' and each 'less than' sign with a 'less than or equal to' sign. We get
$x \le 1$
or
$x \ge 3$
.