# How do you find the roots, real and imaginary, of #y=(x+4)(x+1)-3x^2+2# using the quadratic formula?

Multiply out the bracket:

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To find the roots of the given quadratic equation ( y = (x+4)(x+1) - 3x^2 + 2 ), we first expand it to get the standard quadratic form ( ax^2 + bx + c ). Then, we can apply the quadratic formula, ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where ( a ), ( b ), and ( c ) are coefficients of the quadratic equation.

Expanding the given equation: ( y = (x+4)(x+1) - 3x^2 + 2 ) ( y = x^2 + x + 4x + 4 - 3x^2 + 2 ) ( y = -2x^2 + 5x + 6 )

Comparing with the standard form ( ax^2 + bx + c ): ( a = -2 ), ( b = 5 ), ( c = 6 )

Now, plug these values into the quadratic formula and solve for ( x ): ( x = \frac{-5 \pm \sqrt{5^2 - 4(-2)(6)}}{2(-2)} ) ( x = \frac{-5 \pm \sqrt{25 + 48}}{-4} ) ( x = \frac{-5 \pm \sqrt{73}}{-4} )

Hence, the roots of the given quadratic equation are: ( x = \frac{-5 + \sqrt{73}}{-4} ) (real) ( x = \frac{-5 - \sqrt{73}}{-4} ) (real)

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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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