How do you solve #x^2 - 14x + 49 = 3#?

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Answer 1 · 2015-10-30T05:02:14

# x= 7±sqrt(3) #

You make the equation equal zero so you subtract three from both sides.

#x^2 -14x+46 = 0#
You can't factor it so you can either complete the square or use the quadratic formula

#x=(-b±sqrt(b^2-4ac))/(2a)#

#a=1, b=-14, c=46#
Insert that and you get

# x=(-(-14)±sqrt((-14)^2-4(1)(46)))/(2(1))#

#x= (14±sqrt(196-184))/(2)#

#x= (14±sqrt(12))/(2)#

Factor #sqrt(12)# to #sqrt(4*3# then square root the 4 to 2 making #2sqrt(3)#

# x= (14±2sqrt(3))/(2) #
Simplify

# x= 7±sqrt(3) #