# How do you solve x^2+8x+18 = 0 by completing the square?

Jun 4, 2016

No Solution

#### Explanation:

To complete the square, we need the perfect square of the equation of ${x}^{2} + 8 x + 18$ In order to find the perfect square, we need to change the equation into ${\left(x - b\right)}^{2} = a$, were a and b are constants. To find c, we divide the coefficient by 2 and square it

${\left(\frac{8}{2}\right)}^{2} = 16$

We get 16, which means that we must change our current equation to have a 16.

${x}^{2} + 8 x + 18 - 2 = - 2$

By subtracting 2 from both sides, we get that 16. Now, we can simplify the left hand side into the perfect square

${x}^{2} + 8 x + 16 = {\left(x + 4\right)}^{2}$

This means ${\left(x + 4\right)}^{2} = - 2$

We now square root both side, giving us $x + 4 = \sqrt{-} 2$

They can never be a negative square root, so therefore there is no answer.