we have
[tex]3x^{2}=7x-3[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]3x^{2}-7x=-3[/tex]
Factor the leading coefficient
[tex]3(x^{2}-(7/3)x)=-3[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]3(x^{2}-(7/3)x+(49/36))=-3+(49/12)[/tex]
[tex]3(x^{2}-(7/3)x+(49/36))=(13/12)[/tex]
Rewrite as perfect squares
[tex]3(x-(7/6))^{2}=(13/12)[/tex]
[tex](x-(7/6))^{2}=(13/36)[/tex]
square root both sides
[tex]x-\frac{7}{6}=(+/-) \sqrt{\frac{13}{36}}[/tex]
[tex]x-\frac{7}{6}=(+/-) \frac{\sqrt{13}}{6}[/tex]
[tex]x=\frac{7}{6}(+/-) \frac{\sqrt{13}}{6}[/tex]
The solutions are
[tex]x=\frac{7}{6}+\frac{\sqrt{13}}{6}=\frac{-7+\sqrt{13}}{6}[/tex]
[tex]x=\frac{7}{6}-\frac{\sqrt{13}}{6}=\frac{-7-\sqrt{13}}{6}[/tex]
therefore
The answer is
[tex]x=\frac{7\pm\sqrt{13}}{6}[/tex]
