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How do you solve #-x ^ { 2} - 4x - 3= 3x + 4#?

How do you solve #-x ^ { 2} - 4x - 3= 3x + 4#?
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Answer #1

#x=(-7+sqrt(21))/2# or #x=(-7-sqrt(21))/2#

Explanation:

Given
#color(white)("XXX")-x^2-4x-3=3x+4#

After re-arranging into standard form:
#color(white)("XXX")x^2+7x+4=0#

The applying the quadratic formula
#color(white)("XXX")x=(-7+-sqrt(7^2-4 * 1 * 7))/(2 * 1) = (-7+-sqrt(21))/2#

answered by: Alan P.
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