# How do you solve 3x^2+12x+81=15 by completing the square?

Jun 30, 2017

$x = - 2 \pm i \sqrt{22}$

#### Explanation:

Note that all of the terms are divisible by $3$, so let's divide both sides of the equation by $3$ to get:

${x}^{2} + 4 x + 27 = 5$

To make the left hand side into a perfect square, subtract $23$ from both sides to get:

${x}^{2} + 4 x + 4 = - 22$

The left hand side is now ${\left(x + 2\right)}^{2}$, so our equation is:

${\left(x + 2\right)}^{2} = - 22$

The square of any real number is non-negative, so this quadratic equation only has non-real complex solutions.

If you want to proceed further, note that if $n > 0$ then:

$\sqrt{- n} = i \sqrt{n}$

where $i$ is the imaginary unit, satisfying ${i}^{2} = - 1$.

So we find:

${\left(x + 2\right)}^{2} = {\left(i \sqrt{22}\right)}^{2}$

and hence:

$x + 2 = \pm i \sqrt{22}$

Subtracting $2$ from both sides we get:

$x = - 2 \pm i \sqrt{22}$