# How do you solve n^ { 2} - 20n - 96= 0?

Jan 8, 2018

Factorise and then solve by setting the brackets = to 0.

See below.

#### Explanation:

To solve a quadratic we know that we must get factors of $c$ that add/ subtract to give $b$

$\therefore$ we need factors of $- 96$ that add/ subtract to give $- 20$

After playing around with the numbers you should find that $- 24$ and $4$ work perfectly.

$\therefore {n}^{2} - 20 n - 6 = 0$
is the same as $\left(n - 24\right) \left(n + 4\right) = 0$

We know that a quadratic will always $= 0$, meaning one of our brackets must = 0.

$\therefore \left(n - 24\right) \left(n + 4\right) = 0$

$n = 24 , n = - 4$