# How do you solve -10n^2-10=470?

Jan 26, 2017

$n = \pm 4 i \sqrt{3}$

#### Explanation:

If we start with $- 10 {n}^{2} - 10 = 470$, our first step is to isolate the variable ($n$). To do that, we add $10$ to both sides, giving us: $- 10 {n}^{2} = 480$. If we divide by -10 on both sides we get n^2=-48. We want to solve for $n$, not ${n}^{2}$, so we need to "undo" the square by square-rooting it. What we do on one side, we must do to the other side, so we have $\sqrt{{n}^{2}} = \pm \sqrt{- 48}$ or $n = \pm \sqrt{- 48}$.

We can rewrite $\sqrt{- 48}$ as $\sqrt{- 1} \cdot \sqrt{48}$. $\sqrt{- 1}$ is $i$, and $\sqrt{48}$ can be simplified to $4 \sqrt{3}$.

Putting it all together, we get $n = \pm 4 i \sqrt{3}$