# How do you use the quadratic formula to solve #x^2+4x-2=0#?

See a solution process below:

According to the quadratic formula,

Replacing:

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To solve the equation (x^2 + 4x - 2 = 0) using the quadratic formula, which is (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), identify (a), (b), and (c) from the equation (ax^2 + bx + c = 0).

In this equation, (a = 1), (b = 4), and (c = -2).

Substitute these values into the quadratic formula: [x = \frac{-(4) \pm \sqrt{(4)^2 - 4(1)(-2)}}{2(1)}]

Simplify inside the square root: [x = \frac{-4 \pm \sqrt{16 + 8}}{2}] [x = \frac{-4 \pm \sqrt{24}}{2}]

Further simplify the square root: [x = \frac{-4 \pm 2\sqrt{6}}{2}]

Now, simplify the expression by dividing each term by 2: [x_1 = \frac{-4 + 2\sqrt{6}}{2} = -2 + \sqrt{6}] [x_2 = \frac{-4 - 2\sqrt{6}}{2} = -2 - \sqrt{6}]

So, the solutions to the equation (x^2 + 4x - 2 = 0) using the quadratic formula are (x = -2 + \sqrt{6}) and (x = -2 - \sqrt{6}).

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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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