How do you solve #3y²+14y=5#?

1 Answer
Mar 27, 2016

Answer:

#1/3 and -5#

Explanation:

#f(y) = 3y^2 + 14y = 5 = 0.#
Use the new Transforming Method (Google, Yahoo Search).
Transformed equation: #f'(y) = y^2 + 14y - 15.#
Since a + b + c = 0, the 2 real roots of f'(y) are: ( 1) and #(c/a = - 15)#.
Therefor, the 2 real roots of f(y) are: #(1/3) and (-15/3 = - 5)#
Answers: #1/3# and -5