How do you solve for x in simplest radical form: #2(x+3)^2+10=66#?

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Answer 1 · 2015-04-14T18:04:09

#2(x+3)^2+10=66#

#(x+3)^2 + 5 = 33#

#x^2+6x+9+5-33=0#

#x^2+6x-19=0#

Using the quadratic root formula: #(-b+-sqrt(b^2-4ac))/2a#

#x=(-6+-sqrt(36+76))/2#

#x= (-6+-4sqrt(7))/2#

#x=-3-2sqrt(7)#
or
#x=-3+2sqrt(7)#

Answer 2 · 2015-04-14T22:06:12

Alternative solution:

#2(x+3)^2 +10 = 66#

#2(x+3)^2 = 66 -10# #color(white)"ss"# added #-10# on both sides

#2(x+3)^2 = 56#

#(x+3)^2 = 56/2##color(white)"ssssssss"# multiplied #1/2# on both sides

#(x+3)^2 = 28#

#x+3 = +- sqrt 28# #color(white)"ssssss"# there are 2 numbers whise square is 28

#x = -3 +- sqrt 28# #color(white)"ss"# added #-3# on both sides

#x = -3 +- sqrt(4*7) = -3 +- sqrt4 sqrt7 = -3 +- 2sqrt7#