# How do you solve using the quadratic formula for x^2 + x + 5 = 0?

Apr 13, 2018

The answer is $\frac{- 1 \pm i \sqrt{19}}{2}$.

#### Explanation:

The quadratic formula is x=(-b+-sqrt(b^2-4ac))/(2a for the equation $a {x}^{2} + b x + c$.

In this case, $a = 1$, $b = 1$, and $c = 5$.

You can therefore substitute in those values to get:

(-1+-sqrt(1^2-4(1)(5)))/(2(1).

Simplify to get $\frac{- 1 \pm \sqrt{- 19}}{2}$.

Because $\sqrt{- 19}$ is not a real number, we have to stick to imaginary solutions. (If this problem asks for real number solutions, there are none.)

The imaginary number $i$ equals $\sqrt{- 1}$, therefore we can substitute it in:

$\frac{- 1 \pm \sqrt{- 1 \cdot 19}}{2} \rightarrow \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{19}}{2} \rightarrow \frac{- 1 \pm i \sqrt{19}}{2}$, the final answer.

Hope this helps!

Apr 13, 2018

See application of the quadratic formula below in obtaining the result:
$\textcolor{w h i t e}{\text{XXX}} x = - \frac{1}{2} \pm \sqrt{19} i$

#### Explanation:

${x}^{2} + x + 5 = 0$ is equivalent to $\textcolor{red}{1} {x}^{2} + \textcolor{b l u e}{1} x + \textcolor{m a \ge n t a}{5} = 0$

Applying the general quadratic formula x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(magenta)c))/(2color(red)a
for $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{m a \ge n t a}{c} = 0$

to this specific case, we have
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{{\textcolor{b l u e}{1}}^{2} - 4 \cdot \textcolor{red}{1} \cdot \textcolor{m a \ge n t a}{5}}}{2 \cdot \textcolor{red}{1}}$

$\textcolor{w h i t e}{\text{XXXXX}} = \frac{- 1 \pm \sqrt{- 19}}{2}$

There are no Real solutions, but as Complex values:
$\textcolor{w h i t e}{\text{XXX")x=-1/2+sqrt(19)icolor(white)("XXX")"or"color(white)("XXX}} x = - \frac{1}{2} - \sqrt{19} i$