How do you find the zeros of #y=12x^2+8x-15#?

Answer 1

Two real roots: #5/6 and - 3/2#

Solve y = 12x^2 + 8x - 15 = 0 by the new Transforming Method (Socratic Search) Transformed equation y' = x^2 + 8x - 180 = 0 Compose factor pairs of (-180) --> ... (10, -18). This sum is -8 = -b. The 2 real roots of y' are: 10 and - 18. The 2 real roots of y are: #x1 = 10/a = 10/12 = 5/6# and #x2 = - 18/a = -18/12 = -3/2#
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Answer 2

To find the zeros of (y = 12x^2 + 8x - 15), you can use the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 12), (b = 8), and (c = -15). Substituting these values into the formula, you get: (x = \frac{{-8 \pm \sqrt{{8^2 - 4 \cdot 12 \cdot (-15)}}}}{{2 \cdot 12}}). Solving this equation will give you the zeros of the quadratic equation.

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Answer from HIX Tutor

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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