# How do you solve #3x^2+14x+15=0#?

where,

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I would use the quadratic formula.

Equations of the following type can be solved using the quadratic formula:

And appears as follows:

In our instance:

So:

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To solve the quadratic equation 3x^2 + 14x + 15 = 0, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 3, b = 14, and c = 15. Plugging in these values, you get x = (-14 ± √(14^2 - 4(3)(15))) / (2*3). Simplifying further, you get x = (-14 ± √(196 - 180)) / 6, which becomes x = (-14 ± √16) / 6. This simplifies to x = (-14 ± 4) / 6. Therefore, the solutions are x = (-14 + 4) / 6 = -10 / 6 = -5/3 and x = (-14 - 4) / 6 = -18 / 6 = -3. So, the solutions are x = -5/3 and x = -3.

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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.

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