# How do you solve n^2-3n+2=0?

Jun 22, 2017

$n = 1$ and $n = 2$

#### Explanation:

Rewrite the problem:

${n}^{2} - 2 n - 1 n + 2 = 0$

We can see now that this become:

$\left(n - 2\right) \left(n - 1\right) = 0$

Now we set each of the factors equal to $0$ and solve for $n$:

For the first one:

$n - 2 = 0$

$n - 2 + 2 = 0 + 2$

$n = 2$.

For the second one:

$n - 1 = 0$

$n - 1 + 1 = 0 + 1$

$n = 1$

So our set of solutions is:

$n = 1$ and $n = 2$