How do you solve #4x^2-12x=7#?

1 Answer
Oct 7, 2016

#-1/2 and 7/2#

Explanation:

Bring to standard form:
#y = 4x^2 - 12x - 7 = 0.#
Solve by the new Transforming Method (Socratic Search)
Transformed equation: #y' = x^2 - 12x - 14 = 0#
Find 2 numbers (real roots) knowing sum (-b = 12) and product (c = -28). They are (-2, 14).
Back to original y, the 2 real roots are: #x1 = -2/4 = - 1/2# and
#x2 = 14/4 = 7/2#