Question

Which of the following is a solution of 3x2 = 7x - 3?

Answer 1

3x^2 = 7x - 3
3x^2 - 7x = -3
x^2 - 7/3x = -1
x^2 - 7/3x + 49/36 = -1 + 49/36 = 13/36
(x - 7/6)^2 = 13/36
x - 7/6 = + or - sqrt(13)/6
x = 7/6 + or - sqrt(13)/6
x = (7 + or - sqrt(13))/6

Answer 2

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

$x=\frac{7\pm\sqrt{13}}{6}$