# How do you find all the real and complex roots of x^2 + 10x + 26 = 0?

Jul 19, 2018

$x = - 5 \pm i$

#### Explanation:

We can complete the square then use the difference of squares identity:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

with $A = \left(x + 5\right)$ and $B = i$ as follows:

$0 = {x}^{2} + 10 x + 26$

$\textcolor{w h i t e}{0} = {x}^{2} + 2 \left(x\right) \left(5\right) + {\left(5\right)}^{2} + 1$

$\textcolor{w h i t e}{0} = {\left(x + 5\right)}^{2} - {i}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x + 5\right) - i\right) \left(\left(x + 5\right) + i\right)$

$\textcolor{w h i t e}{0} = \left(x + 5 - i\right) \left(x + 5 + i\right)$

Hence:

$x = - 5 \pm i$