# How do you solve x^2 + 5x + 7 = 0 using the quadratic formula?

##### 3 Answers
Mar 22, 2018

$\frac{- 5 + i \sqrt{3}}{2}$ and $\frac{- 5 - i \sqrt{3}}{2}$

#### Explanation:

For quadratic equations of the form:

$a {x}^{2} + b x + c$

The quadratic formula is given by:

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

From given equation we have:

$\boldsymbol{a} = 1$

$\boldsymbol{b} = 5$

$\boldsymbol{c} = 7$

Putting these values in the quadratic formula:

$\frac{- \left(5\right) \pm \sqrt{{\left(5\right)}^{2} - 4 \left(1\right) \left(7\right)}}{2 \left(1\right)} = \frac{- 5 \pm \sqrt{25 - \left(28\right)}}{2}$

$= \frac{- 5 \pm \sqrt{- 3}}{2}$

We can write this in the following way:

$\sqrt{- 3} = \sqrt{3 \times - 1} = \sqrt{3} \cdot \sqrt{- 1}$

If $\sqrt{- 1} = i$

Then:

$\frac{- 5 + i \sqrt{3}}{2}$ and $\frac{- 5 - i \sqrt{3}}{2}$

These are known as complex roots.

Mar 22, 2018

$x = \frac{- 5 + i \sqrt{3}}{2}$ or $\frac{- 5 - i \sqrt{3}}{2}$

#### Explanation:

Accordng to quadratic formula, solution of quadratic equation $a {x}^{2} + b x + c = 0$ is

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Hence solution of ${x}^{2} + 5 x + 7 = 0$ is

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \cdot 1 \cdot 7}}{2}$

= $\frac{- 5 \pm \sqrt{25 - 28}}{2}$

= $\frac{- 5 \pm \sqrt{- 3}}{2}$

i.e. $x = \frac{- 5 + i \sqrt{3}}{2}$ or $\frac{- 5 - i \sqrt{3}}{2}$

Mar 22, 2018

No real solutions

#### Explanation:

${x}^{2} + 5 x + 7 = 0$

Use the quadratic formula with $a = 1 , b = 5 , c = 7$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 5 \pm \sqrt{{\left(5\right)}^{2} - 4 \left(1\right) \left(7\right)}}{\left(2\right) \left(1\right)}$

$x = \frac{- 5 \pm \sqrt{25 - 28}}{2}$

$x = \frac{- 5 \pm \sqrt{- 3}}{2}$

No real solution