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