# How do you solve the equation 3x^2-24x+27=0 by completing the square?

Jul 4, 2017

You may start by dividing everything by $3$

#### Explanation:

$\to {x}^{2} - 8 x + 9 = 0$

You take half of the $x$ coefficient and square that. The square would then be:
${x}^{2} - 8 x + {\left(- 4\right)}^{2} = {x}^{2} - 8 x + 16 = {\left(x - 4\right)}^{2}$

To get there you would have added $7$ to the original $9$. Also add $7$ to the other side of the $=$ sign:

$\to {\left(x - 4\right)}^{2} = 7 \to x - 4 = \pm \sqrt{7} \to x = 4 \pm \sqrt{7}$