# How do you solve 5x^2+64=0?

Oct 20, 2016

This can be solved using the quadratic formula.

#### Explanation:

While there are many ways to solve quadratic equations like this one, the quadratic formula is one of the simplest.

The quadratic formula looks like this:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2} a$

To use this equation, we simply plug in the values given in the problem. To find these values, we must acknowledge that a quadratic equation looks like this:
$y = a {x}^{2} + b x + c$

Once we know this, we can determine that our equation has an a value of 5, because it is multiplied by ${x}^{2}$. The b value must be zero, because none is given. If a number (x) is multiplied by zero, it will disappear. Finally, the c value must be 64, because it is not multiplied by any x value.
Now, we can plug in the values of a, b, and c.
$x = \frac{- 0 \pm \sqrt{{0}^{2} - 4 \left(5\right) \left(64\right)}}{2 \left(5\right)}$

We simplify to get:
$x = \frac{\pm \sqrt{- 1280}}{10}$

The $\pm$ gives us two values for x. We solve for $+$ and $-$ separately. Because the value under the radical is negative, the answer will involve $i$, which is the variable given to the $\sqrt{- 1}$.
When we simplify, the two x values are:
$x = - \frac{8 \sqrt{5} i}{5}$
$x = \frac{8 \sqrt{5} i}{5}$