How do you solve #13y+42=-y^2#?

1 Answer
Nghi N
Apr 5, 2018

-6 and - 7

Explanation:

#y^2 + 13y + 42 = 0#
Use the new Transforming Method (Socratic, Google Search) - Case a = 1.
Find 2 real roots, both negative, knowing their sum (-b = -13), and their product (c = 42).
They are - 6 and - 7

Related questions
Impact of this question
692 views around the world
You can reuse this answer
Creative Commons License