How do you solve #x^2 = -5x - 5#?

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How do you solve #x^2 = -5x - 5#?

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Answer 1 · 2016-06-10T03:07:08

#x = (-5 +- sqrt5)/2#

Explanation:

First, bring the equation to standard form:
#y = x^2 + 5x + 5 = 0#
Use the improved quadratic formula (Google Search)
#D = d^2 = b^2 - 4ac = 25 - 20 = 5# --> #d = +- sqrt5#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = -5/2 +- sqrt5/2 = (-5 +- sqrt5)/2#

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How do you solve #x^2 = -5x - 5#?

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