SOLUTION: Find the solution sets for 2x^2+3x-5=0
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Question 89207
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Find the solution sets for
2x^2+3x-5=0
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2x^2 + 3x - 5 = 0
2x^2 + 5x - 2x - 5 = 0 ... being 5 and -2 are factors of 2*-5 that sum to 3
(2x^2 + 5x) + (-2x - 5) = 0
x(2x + 5) - (2x + 5) = 0
(x - 1)(2x + 5) = 0
x = 1 and x = -5/2