How do you solve #2d ^ { 2} + 29d - 15= 0#?

1 Answer
Jul 21, 2017

Answer:

#1/2# and - 15

Explanation:

Use the new Transforming Method (Google Search):
#f(d) = 2d^2 + 29d - 15 = 0#
Transformed equation:
#f'(d) = d^2 + 29d - 30 = 0#
Proceeding: Find 2 real roots of f'(d), then, divide them by a = 2.
Since a + b + c = 0, use shortcut. The 2 real roots are: 1 and
#c/a = -30#.
Back to f(d) --> The 2 real roots are: #x1 = 1/(a) = 1/2# and
#x2 = -30/a = -30/2 = - 15#