How do you solve #-10n^2-10=470#?

1 Answer
Jan 26, 2017

Answer:

#n=+-4isqrt3#

Explanation:

If we start with #-10n^2-10=470#, our first step is to isolate the variable (#n#). To do that, we add #10# to both sides, giving us: #-10n^2=480#. If we divide by -10 on both sides we get n^2=-48. We want to solve for #n#, not #n^2#, so we need to "undo" the square by square-rooting it. What we do on one side, we must do to the other side, so we have #sqrt(n^2)=+-sqrt(-48)# or #n=+-sqrt(-48)#.

We can rewrite #sqrt(-48)# as #sqrt(-1)*sqrt(48)#. #sqrt(-1)# is #i#, and #sqrt48# can be simplified to #4sqrt3#.

Putting it all together, we get #n=+-4isqrt3#