How do you solve #x^2 - 12x + 80 = 8#?

Given #x^2-12x+80 = 8#

Subtracting #8# from both sides
#color(white)"XXXX"##x^2-12x+72=0#

Using the quadratic formula (see below if you are uncertain of this)
#color(white)"XXXX"##x= (-(-12)+-sqrt((-12)^2-4(1)(72)))/(2(1)#

#color(white)"XXXXXXXX"##=(12 +-sqrt(144 - 288)/2#

#color(white)"XXXXXXXX"##= (12+-sqrt(-144))/2#

#color(white)"XXXXXXXX"##= (12+-12i)/2#

#color(white)"XXXXXXXX"##=6+-6i#

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

#color(white)"XXXXXX"#The need to use this formula comes up often enough that it is worth memorizing

#x^2−12x+80=8#
#x^2−12x+80 -8=0#

#x^2−12x+72=0# .

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=-12, c=72#

The Discriminant is given by:
#Delta=b^2-4*a*c#

# = (-12)^2-(4*1*72)#
# = 144-288=-144#

The solutions are found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

#x = (-(-12)+-sqrt(-144))/(2*1) = (12+-sqrt(-144))/2#

#x=(12+-12i)/2#

#= (2(6+-6i))/2#

#= 6+-6i#

#color(blue)(x=6(1+i)#
#color(blue)( x=6(1-i)#