# How do you solve #x^2 + 5x + 6 = 0# algebraically?

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To solve the quadratic equation (x^2 + 5x + 6 = 0) algebraically, you can use the quadratic formula, which is (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a), (b), and (c) are the coefficients of the quadratic equation (ax^2 + bx + c = 0).

For the given equation (x^2 + 5x + 6 = 0), (a = 1), (b = 5), and (c = 6).

Substitute these values into the quadratic formula:

(x = \frac{{-5 \pm \sqrt{{5^2 - 4(1)(6)}}}}{{2(1)}})

Solve the expression under the square root:

(x = \frac{{-5 \pm \sqrt{{25 - 24}}}}{2})

(x = \frac{{-5 \pm \sqrt{1}}}{2})

(x = \frac{{-5 \pm 1}}{2})

This gives two possible solutions:

- (x = \frac{{-5 + 1}}{2} = \frac{-4}{2} = -2)
- (x = \frac{{-5 - 1}}{2} = \frac{-6}{2} = -3)

So, the solutions to the equation (x^2 + 5x + 6 = 0) are (x = -2) 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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