How do you solve #(x - 4)(x + 5) = 7# using the quadratic formula?
- Simplify the equation by FOIL METHOD
#(x-4)(x+5)=7#
#x^2+5x-4x-20=7#
#x^2+x-27=0 - Identify the values for a, b, and c
#a=1#
#b=1#
#c=-27# - Use the standard formula, which is
#x=(-b+-sqrt(b^2-4ac))/(2a)# - Plug in the values
#x=(-1+-sqrt(1^2-4(1)(-27)))/(2(1))#
#x=(-1+-sqrt(1+108))/2#
#x=(-1+-sqrt109)/2# - Answer
#x=(-1+-sqrt109)/2#
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To solve the equation (x - 4)(x + 5) = 7 using the quadratic formula, first, expand the expression (x - 4)(x + 5) to get x^2 + x - 20. Then, rewrite the equation as x^2 + x - 20 = 7. Next, rearrange terms to get x^2 + x - 27 = 0. Now, apply the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = 1, and c = -27. Substitute these values into the formula and solve for x.
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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.
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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