How do you solve # 2x^2 + 3x - 14 = 0#?

1 Answer
Nghi N
Jun 30, 2017

2, and #- 7/2#

Explanation:

#2x^2 + 3x - 14 = 0#
Use new Transforming Method.
Transformed equation:
#y' = x^2 + 3x - 28 = 0#
Proceeding: find 2 real roots of y', then, divide them by a = 2.
Find 2 real roots, knowing sum (-b = -3) and product (ac = - 28). They are: 4 and -7 (because sum = -3 and product = - 28).
Back to original y, the 2 real roots are:
#x1 = 4/a = 4/2 = 2#, and #x2 = - 7/a = - 7/2 = - 7/2#

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