Answer:
x < -5/2 or 3 < x
Explanation:
Subtracting 15 puts the relation into a more standard form:
2x^2 -x -15 > 0
(2x +5)(x -3) > 0 . . . . . factor
The zeros of these factors are x = -5/2 and x = 3.
For values of x between those values (-5/2 ≤ x ≤ 3), the product of these factors will be negative or zero. In order for the product to be positive, x must not be in that range.
The solution is ...
x < -5/2 or 3 < x
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Additional comment
Several methods are taught for factoring a quadratic expression. One of my favorite is using a graphing calculator to find the zeros. If p is a zero, then (x-p) is a factor.
Another way to approach the factoring of ax^2 +bx +c is to find factor pairs of the product (ac) that have a total of b. Here, that means finding factors of 2(-15) = -30 that have a sum of -1. Consider ...
-30 = (1)(-30) = 2(-15) = 3(-10) = 5(-6)
The sums of these pairs are -29, -13, -7, -1. With the values 5 and -6, you can rewrite the quadratic expression as 2x^2 +5x -6x -15. This can be factored by grouping pairs of terms:
x(2x +5) -3(2x +5)
(2x +5)(x -3) . . . . . . . the factorization we show above.
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You can see on the graph that the quadratic is greater than 0 when x < -2.5 or x > 3.