Answer
The solution set is\[\left\{ 1,3 \right\}\].
Work Step by Step
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\]
Here,\[a=1,\,\,b=-4,\text{ and }c=3\].
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is:
\[\begin{align}
& x=\frac{4\pm \sqrt{{{\left( -4 \right)}^{2}}-4\times 1\times 3}}{2\times 1} \\
& =\frac{4\pm \sqrt{16-12}}{2} \\
& =\frac{4\pm \sqrt{4}}{2} \\
& =\frac{4\pm 2}{2}
\end{align}\]
Further simplify it to get:
\[\begin{align}
& x=\frac{4\pm 2}{2} \\
& =\frac{4-2}{2},\frac{4+2}{2} \\
& =\frac{2}{2},\frac{6}{2} \\
& =1,3
\end{align}\]
