How do you solve #2x^2-5x-3=0#?

1 Answer
Noah G
Jan 8, 2016

It would be simplest to solve by factoring.

Explanation:

#2x^2# - 5x - 3 = 0

#2x^2# - 6x + x - 3 = 0

2x(x - 3) + 1(x - 3) = 0

(2x + 1)(x - 3) = 0

x = #-1/2# and 3

The solutions to your equation are x = #-1/2# and x = 3.

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