# What kind of solutions does 5x^2 + 8x + 7 = 0 have?

Jun 15, 2018

#### Explanation:

The key realization is that our quadratic is of the form

$a {x}^{2} + b x + c = 0$, where the value ${b}^{2} - 4 a c$ is the discriminant, which tells us

$I . {b}^{2} - 4 a c > 0 \implies 2$ real roots

$I I . {b}^{2} - 4 a c < 0 \implies$ no real roots

$I I I . {b}^{2} - 4 a c = 0 \implies 1$ real root

In our quadratic, we know $a = 5 , b = 8$ and $c = 7$. So let's plug these values in. We get

${8}^{2} - 4 \left(5 \cdot 7\right)$

$\implies 64 - 4 \left(35\right)$

$64 - 140 = \textcolor{b l u e}{- 76}$

Notice, our discriminant is less than zero, so we are in scenario $I I$. This means our quadratic has no real solutions.

Hope this helps!

The roots are all complex numbers.

#### Explanation:

The quadratic equation is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Here, $a = 5$, $b = 8$, and $c = 7$.

Substituting, we get $x = \frac{- 8 \pm \sqrt{{8}^{2} - 4 \cdot 5 \cdot 7}}{2 \cdot 5}$.

Simplify.

$x = \frac{- 8 \pm \sqrt{64 - 140}}{10}$

Here, we can already tell that the answers will be complex numbers. If you want to simplify further...

$x = \frac{- 8 \pm \sqrt{- 76}}{10}$

$x = \frac{- 8 \pm 4 \sqrt{- 19}}{10}$

$x = \frac{- 4 \pm 2 \sqrt{- 19}}{5}$