How do you solve #x^2-2x-2=0#?

1 Answer
Aug 6, 2015

Answer:

Use the quadratic formula to get
#color(white)("XXXX")##x= 1+-sqrt(3)#

Explanation:

For any quadratic equation of the form
#color(white)("XXXX")##ax^2+bx+c=0#
the solutions are given by the formula
#color(white)("XXXX")##x= (-b+-sqrt(b^2-4ac))/(2a)#

Given #x^2-2x-2=0#
#a=1##color(white)("XXXX")##b=-2##color(white)("XXXX")##c=-2#
and the quadratic formula gives
#color(white)("XXXX")##x= (2+-sqrt((-2)^2-4(1)(-2)))/(2(1))#

#color(white)("XXXX")##color(white)("XXXX")##= (2+-sqrt(12))/2#

#color(white)("XXXX")##color(white)("XXXX")##=(2+-2sqrt(3))/2#

#color(white)("XXXX")##color(white)("XXXX")##=1+-sqrt(3)#