# How do you solve 4w^2+100=0 using the quadratic formula?

Apr 27, 2017

There are no real solutions.
$w = - 5 i$
$w = 5 i$

#### Explanation:

The quadratic formula is as follows
$\setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$
Given a function in the form
$a {w}^{2} + b w + c = 0$

Plugging in the values we have, we the following
$\setminus \frac{- \left(0\right) + \setminus \sqrt{{\left(0\right)}^{2} - 4 \left(4\right) \left(100\right)}}{2 \left(4\right)}$
$\setminus \frac{- \left(0\right) - \setminus \sqrt{{\left(0\right)}^{2} - 4 \left(4\right) \left(100\right)}}{2 \left(4\right)}$

Solving for the first one we get
$\setminus \frac{\setminus \sqrt{0 - 1600}}{8} = \setminus \frac{\setminus \sqrt{- 1600}}{8}$

Solving for the second one we get
$\setminus \frac{- \setminus \sqrt{0 - 1600}}{8} = \setminus \frac{- \setminus \sqrt{- 1600}}{8}$

There are no real solutions, but we can get the imaginary/complex solutions
$\setminus \frac{\setminus \sqrt{1600 \setminus \cdot - 1}}{8} = \setminus \frac{40 i}{8} = 5 i$
$\setminus \frac{- \setminus \sqrt{1600 \setminus \cdot - 1}}{8} = \setminus \frac{- 40 i}{8} = - 5 i$