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

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Answer 1 · 2017-06-22T20:10:46

#n=1# and #n=2#

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#