# How do you find the roots, real and imaginary, of #y=-2(x +3)^2-5x + 6# using the quadratic formula?

Roots:

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To find the roots of the quadratic equation (y = -2(x + 3)^2 - 5x + 6), we first need to rewrite it in the standard form (ax^2 + bx + c = 0). After that, we can use the quadratic formula, which states that for an equation of the form (ax^2 + bx + c = 0), the roots are given by (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}).

In this case, we have (y = -2(x + 3)^2 - 5x + 6). Expanding and simplifying, we get (y = -2(x^2 + 6x + 9) - 5x + 6), which further simplifies to (y = -2x^2 - 12x - 18 - 5x + 6), or (y = -2x^2 - 17x - 12).

Comparing this to the standard form (ax^2 + bx + c = 0), we have (a = -2), (b = -17), and (c = -12).

Plugging these values into the quadratic formula, we get:

[x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(-2)(-12)}}{2(-2)}]

Simplifying further:

[x = \frac{17 \pm \sqrt{289 - 96}}{-4}]

[x = \frac{17 \pm \sqrt{193}}{-4}]

So, the roots are:

[x = \frac{17 + \sqrt{193}}{-4}]

[x = \frac{17 - \sqrt{193}}{-4}]

These are the real and imaginary roots of the given quadratic equation.

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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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