# How do you use the quadratic formula to solve #9x^2-6x-35=0#?

Given:

Be aware that this is written in standard form:

The quadratic formula yields the roots:

That is:

Although the quadratic formula is very helpful and worthy of memorization, if you haven't already, I highly advise you to learn how to derive it from scratch.

Using the identity of the difference of squares, here's one method:

Given:

We are able to write:

Hence:

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To use the quadratic formula to solve the equation (9x^2 - 6x - 35 = 0), we first identify the coefficients (a), (b), and (c) in the general quadratic equation (ax^2 + bx + c = 0). Then, we substitute these values into the quadratic formula:

[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]

For the equation (9x^2 - 6x - 35 = 0), (a = 9), (b = -6), and (c = -35). Substituting these values into the quadratic formula, we get:

[x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4(9)(-35)}}}}{{2(9)}}]

[x = \frac{{6 \pm \sqrt{{36 + 1260}}}}{{18}}]

[x = \frac{{6 \pm \sqrt{{1296}}}}{{18}}]

[x = \frac{{6 \pm 36}}{{18}}]

Now, we have two possible solutions:

[x_1 = \frac{{6 + 36}}{{18}} = \frac{{42}}{{18}} = \frac{{7}}{{3}}]

[x_2 = \frac{{6 - 36}}{{18}} = \frac{{-30}}{{18}} = -\frac{{5}}{{3}}]

So, the solutions to the equation (9x^2 - 6x - 35 = 0) are (x = \frac{{7}}{{3}}) and (x = -\frac{{5}}{{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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