# How do you solve 4x^2+40x+280=0 by completing the square?

Jul 2, 2017

You can't, unless you want to have complex numbers. If you want complex numbers, then the answer would be $x = - 5 \setminus \pm 3 i \setminus \sqrt{5}$

#### Explanation:

You can simplify the quadratic to make completing the square easier:
$4 {x}^{2} + 40 x + 280 = 0$
${x}^{2} + 10 x + 70 = 0$

Completing the square:
${x}^{2} + 10 x + 70 = 0$
${x}^{2} + 10 x = - 70$
${x}^{2} + 10 x + {\left(\frac{10}{2}\right)}^{2} = - 70 + {\left(\frac{10}{2}\right)}^{2}$
${x}^{2} + 10 x + 25 = - 70 + 25$
${\left(x + 5\right)}^{2} = - 45$

We cannot continue here, since there is no way for the above equation to be true.

However, if you want to have complex numbers, we can continue:
$x + 5 = \setminus \pm \setminus \sqrt{- 45}$
$x + 5 = \setminus \pm i \setminus \sqrt{45}$
$x + 5 = \setminus \pm 3 i \setminus \sqrt{5}$
$x = - 5 \setminus \pm 3 i \setminus \sqrt{5}$