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

1 Answer
Jun 22, 2017

Answer:

#n=1# and #n=2#

Explanation:

Rewrite the problem:

#n^2-2n-1n+2=0#

We can see now that this become:

#(n-2)(n-1)=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#