# How do you solve  c^2+6c-27=0 by completing the square?

May 7, 2015

The answer is $c = 3$, $c = - 9$ .

Solve ${c}^{2} + 6 c - 27 = 0$ by completing the square.

Add $27$ to both sides.

${c}^{2} + 6 c = 27$

Halve the coefficient of $6 c$ then square it.
$\frac{6}{2} = 3$

${3}^{2} = 9$

Add $9$ to both sides of the equation.

${c}^{2} + 6 c + 9 = 27 + 9$ =

${c}^{2} + 6 c + 9 = 36$

The lefthand side of the equation is now a perfect trinomial square. Factor the perfect trinomial square. ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$

$a = c$, $b = 3$

${\left(c + 3\right)}^{2} = 36$

Take the square root of both sides and solve for $c$.

$c + 3 = \pm \sqrt{36}$

$c + 3 = \pm 6$

When $c + 3 = 6$:

$c = 6 - 3$, $c = 3$

When $c + 3 = - 6$:

$c = - 6 - 3$, $c = - 9$

Check:

${3}^{2} + 6 \cdot 3 - 27 = 0$ =

$9 + 18 - 27 = 0$ =

$27 - 27 = 0$

$- {9}^{2} + 6 \left(- 9\right) - 27 = 0$ =

$81 - 54 - 27 = 0$ =

$81 - 81 = 0$