How do you solve #(3x - 8) ^ { 2} - 6= 58#?

#(3*x − 8)^2 − 6 = 58# ; #9*x^2 − 48*x + 64 − 6 = 58# ; #9*x^2 − 48*x + 64 − 6 - 58 = 0# ;

#9*x^2 − 48*x + 0 = 0#

We now have the equation in the form of a quadratic, and can solve it by using the quadratic formula. See http://www.purplemath.com/modules/quadform.htm for instructions.

Algebraically we could also divide by x to obtain #9*x − 48 = 0# ; #9*x = 48# ; #x = 5.33#
CHECK: #(3*5.33 − 8)^2 − 6 = 58# ; #(16 − 8)^2 − 6 = 58# : #(8)^2 − 6 = 58# : #64 − 6 = 58# ; # 58 = 58# CORRECT
This is ONLY valid IF x is NOT '0' (which is one solution in this case). Using the Quadratic Formula or the simplification method used in the other answer is preferred.

My original alternative was just an example of not using the quadratic formula, and it has inadvertently, but helpfully, highlighted the error of the 'short cut".

This equation is in the form #a^2 = b#
We can solve it by finding the square root of both sides.

#(3x-8)^2 -6 = 58#

#(3x-8)^2 = 64#

#3x-8 = +-sqrt64#

First option:

#3x -8 = +8#

#3x =16#

#x = 16/3#

Second option:

#3x-8 = -8#

#3x = 0#

#x=0#

Check #x = 16/3color(white)(..................................)x=0#:

#(3(16/3) -8)^2-6color(white)(...................)(3(0)-8)^2 -6#

#=8^2 -6color(white)(..................................)=(-8)^2 -6#

#=58color(white)(.........................................)=58#