# How do you solve 9y^2+16= -24?

Jun 25, 2015

I am not sure that your question is complete so I included two versions: have a look:

#### Explanation:

1] As it is:
$9 {y}^{2} + 16 = - 24$
$9 {y}^{2} = - 16 - 24$
${y}^{2} = - \frac{40}{9}$
If you use complex numbers you can introduce the imaaginary unit $i = \sqrt{- 1}$ and get:
$y = \pm \sqrt{- \frac{40}{9}}$
$y = \pm \sqrt{- 1 \cdot \frac{40}{9}}$
$y = \pm i \sqrt{\frac{40}{9}} = \pm i \sqrt{4 \cdot \frac{10}{9}} = \pm \frac{2}{3} i \sqrt{10}$

2] How I think it is:
$9 {y}^{2} + 16 = - 24 \textcolor{red}{x}$
$9 {y}^{2} + 24 y + 16 = 0$
${y}_{1 , 2} = \frac{- 24 \pm \sqrt{576 - 576}}{18} = - \frac{24}{18} = - \frac{4}{3}$