How do you solve #n^2-120=7n#?

2 Answers
Apr 1, 2018

Answer:

#n=15,-8#

Explanation:

Whenever we're dealing with quadratics, we want to set them equal to zero, to find the zeroes.

We can start by subtracting #7n# from both sides to get:

#n^2-7n-120=0#

Now, we need to think of two numbers that sum up to #-7# and their product is #-120#. Since their product is negative, the signs are different.

After some trial and error, we arrive at #-15# and #8#, as

#-15+8=-7#

and

#-15*8=-120#

Now, we have the following:

#(n-15)(n+8)=0#

To find the zeroes, we just take the inverse signs to get:

#n=15# & #n=-8#

Hope this helps!

Apr 1, 2018

Answer:

-8 and 15.

Explanation:

#n^2 - 7n - 120 = 0#.
Solve this equation by the new Transforming Method (Socratic, Google Search), case a = 1.
Find 2 real roost, that have opposite signs (ac < 0), knowing their sum (- b = 7) and their product (c = -120).
They are: - 8 and 15.

Note . This method avoids the lengthy factoring by grouping and solving the 2 binomials.