# How do you solve n^2-5=-4?

Sep 1, 2016

$n = \pm 1$

#### Explanation:

To solve a quadratic equation we require to equate it to zero.

The first step is therefore to add 4 to both sides of the equation.

$\Rightarrow {n}^{2} - 5 + 4 = \cancel{- 4} + \cancel{4} = 0$

$\Rightarrow {n}^{2} - 1 = 0 \text{ is the equation to be solved}$

Now ${n}^{2} - 1$ is a $\textcolor{b l u e}{\text{difference of squares}}$

$\Rightarrow \left(n - 1\right) \left(n + 1\right) = 0$

solve: $n - 1 = 0 \Rightarrow n = 1$

solve $n + 1 = 0 \Rightarrow n = - 1$

Thus the solutions to the equation are $n = \pm 1$

Sep 4, 2016

$n = \pm 1$

#### Explanation:

Although this is a quadratic equation, it is a special case because there is no 'n' term.

Isolate the ${n}^{2}$ term.

${n}^{2} = - 4 + 5$

${n}^{2} = 1$

$n = \pm 1$