Answer
The solution of the given inequality is $x\in \left( -\infty ,1 \right)\cup \left( 3,\infty \right)$.
Work Step by Step
The inequality can be solved as follows:
$\begin{align}
& {{x}^{2}}-4x>-3 \\
& {{x}^{2}}-4x+3>0
\end{align}$
Now, apply the method of factorization:
$\begin{align}
& {{x}^{2}}-3x-x+3>0 \\
& x\left( x-3 \right)-1\left( x-3 \right)>0 \\
& \left( x-1 \right)\left( x-3 \right)>0
\end{align}$
The zeros of the above inequality are 1 and 3.
Now, check the different values of x in three regions on the number line system:
For $x<1$ , choose any number in this region; suppose $x=0.5$.
$\left( 0.5-1 \right)\left( 0.5-3 \right)=1.25$
In this region, the result is positive, which means that in this region the graph is positive. Thus, the region $x<1$ is satisfied by the inequality.
For $13$, choose any number in this region; suppose $x=4$.
$\left( 4-1 \right)\left( 4-3 \right)=3$
In this region, the result is positive, which means that in this region the graph is positive. Thus, the region $x>3$ is satisfied by the inequality.
Therefore, the solution is $x\in \left( -\infty ,1 \right)\cup \left( 3,\infty \right)$.
