How do you solve #-30x^{2} - 20x = - 2x^{2} - 4x#?

1 Answer
Nghi N.
Sep 16, 2016

x = 0
#x = -4/7#

Explanation:

Bring the equation to standard form:
#y = - 30x^2 - 20x + 2x^2 + 4x = - 28x^2 - 16x = 0#
#y = x(- 28x - 16) = 0#
There are 2 real roots:
a. x = 0
b. - 28x - 16 = 0
- 28x = 16
#x = - 16/28 = -4/7#

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