# How do you solve 3x^2+14x=5 by completing the square?

May 12, 2015

Dividing through by $3$, we get:

${x}^{2} + \left(\frac{14}{3}\right) x = \frac{5}{3}$

Now try squaring $x + \frac{7}{3}$

${\left(x + \frac{7}{3}\right)}^{2} = \left(x + \frac{7}{3}\right) \left(x + \frac{7}{3}\right)$

$= {x}^{2} + \left(\frac{14}{3}\right) x + \frac{49}{9}$

$= \frac{5}{3} + \frac{49}{9}$

$= \frac{15}{9} + \frac{49}{9}$

$= \frac{64}{9}$

$= \frac{{8}^{2}}{{3}^{2}}$

Hence $x + \frac{7}{3} = \pm \frac{8}{3}$

Subtracting 7/3 from both sides, we get

$x = \pm \frac{8}{3} - \frac{7}{3}$

So $x = \frac{1}{3}$ or $x = - \frac{15}{3} = - 5$.