# How do you solve #2x^2+6x+4=0# using the quadratic formula?

The solutions for the equation are:

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To solve the quadratic equation (2x^2 + 6x + 4 = 0) using the quadratic formula, we first identify the coefficients: (a = 2), (b = 6), and (c = 4). Then, we plug these values into the quadratic formula:

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

Substitute the values of (a), (b), and (c):

[x = \frac{{-6 \pm \sqrt{{6^2 - 4 \cdot 2 \cdot 4}}}}{{2 \cdot 2}}]

Simplify the expression under the square root:

[x = \frac{{-6 \pm \sqrt{{36 - 32}}}}{{4}}]

[x = \frac{{-6 \pm \sqrt{{4}}}}{{4}}]

[x = \frac{{-6 \pm 2}}{{4}}]

Now, we have two possible solutions:

[x_1 = \frac{{-6 + 2}}{{4}} = \frac{{-4}}{{4}} = -1]

[x_2 = \frac{{-6 - 2}}{{4}} = \frac{{-8}}{{4}} = -2]

Therefore, the solutions to the equation (2x^2 + 6x + 4 = 0) are (x = -1) and (x = -2).

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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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