The general formula for a equation of the form:
[tex]ax^2+bx+c=0[/tex]is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]In this case we notice that a=1, b=-4 and c=3. Plugging this values in the general formula we get:
[tex]\begin{gathered} x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(3)}}{2(1)} \\ =\frac{4\pm\sqrt[]{16-12}}{2} \\ =\frac{4\pm\sqrt[]{4}}{2} \\ =\frac{4\pm2}{2} \end{gathered}[/tex]then:
[tex]x_1=\frac{4+2}{2}=\frac{6}{2}=3[/tex]and
[tex]x_2=\frac{4-2}{2}=\frac{2}{2}=1[/tex]Therefore, x=3 or x=1.
