How do you solve #x( 5x + 3) = 2x^{2} - 2x + 28#?

1 Answer
Sep 15, 2016

#-7/3 and 4#

Explanation:

Develop and bring the equation to quadratic standard form:
#y = 3x^2 + 5x - 28 = 0#
Solve it by the new Transforming Method (Socratic Search)
Transformed equation: #y' = x^2 + 5x - 84#.
Roots have opposite signs.
Compose factor pairs of (ac = -84) -->...(-6, 14)(-7, 12). This sum is (12 - 7 = 5 = -b). Then, the 2 real roots of y' are: -7 and 12.
Back to y, its 2 real roots are;
#x1 = -7/a = -7/3#, and #x2 = 12/3 = 4#