How do you find the zeroes of #g(x)=2x^2-5x-3 #?
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To find the zeroes of the function g(x) = 2x^2 - 5x - 3, you can use the quadratic formula, which is given by:
x = [-b ± √(b^2 - 4ac)] / (2a)
where a = 2, b = -5, and c = -3. Substituting these values into the formula, you get:
x = [5 ± √((-5)^2 - 4(2)(-3))] / (2*2)
Simplify the expression inside the square root:
x = [5 ± √(25 + 24)] / 4
x = [5 ± √49] / 4
x = [5 ± 7] / 4
This gives you two possible solutions:
- x = (5 + 7) / 4 = 12 / 4 = 3
- x = (5 - 7) / 4 = -2 / 4 = -0.5
So, the zeroes of the function g(x) = 2x^2 - 5x - 3 are x = 3 and x = -0.5.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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