# How do you solve #x^2+3x+2=0# using the quadratic formula?

To solve the quadratic equation (x^2 + 3x + 2 = 0) using the quadratic formula, where (ax^2 + bx + c = 0), the quadratic formula states that the solutions for (x) are given by:

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

For the given equation (x^2 + 3x + 2 = 0), (a = 1), (b = 3), and (c = 2).

Substituting these values into the quadratic formula:

[x = \frac{{-3 \pm \sqrt{{3^2 - 4(1)(2)}}}}{{2(1)}}]

[x = \frac{{-3 \pm \sqrt{{9 - 8}}}}{{2}}]

[x = \frac{{-3 \pm \sqrt{1}}}{{2}}]

[x = \frac{{-3 \pm 1}}{{2}}]

Thus, the solutions for (x) are (x = \frac{{-3 + 1}}{{2}}) and (x = \frac{{-3 - 1}}{{2}}).

Solving these:

[x_1 = \frac{{-3 + 1}}{{2}} = \frac{{-2}}{{2}} = -1]

[x_2 = \frac{{-3 - 1}}{{2}} = \frac{{-4}}{{2}} = -2]

Therefore, the solutions for the equation (x^2 + 3x + 2 = 0) are (x = -1) and (x = -2).

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The two solutions are

Since this question is given in standard form, meaning that it follows the form:

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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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