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

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Answer 1 · 2015-08-01T18:16:38

#color(red)(x=(sqrt5-5)/2)# and #color(red)(x = -(sqrt5+5)/2)#

#x^2 = -5x-5#

Convert the equation to standard form,

#ax^2 + bx +c =0#

#x^2 + 5x+5 = 0#

#a=1#, #b=5#, and #c=5#

#x = (-b±sqrt(b^2 -4ac))/(2a) = (-5±sqrt(5^2-4×1×5))/(2×1) = (-5±sqrt(25-20))/2 = (-5±sqrt5)/2#

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

or

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