How do you solve #(c - 3) ( c - 1) = 6#?

First we can put the equation into standard form #ax^2+bx+c=0#

#(c-3)(c-1)=6#

#<=># Use FOIL to expand the polynomial

#c^2-4c+3=6#

#<=># Subtract #6# from both sides

#c^2-4c-3=0#

Since you can't factor this polynomial, we can use the quadratic formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case

#a=1#

#b=-4#

#c=-3#

and we replace #x# with the variable #c#

then we plug in

#=> c=(4+-sqrt(16-4(1)(-3)))/2=(4+-sqrt(16+12))/2=(4+-sqrt(28))/2#

Since #sqrt(ab)=sqrt(a)sqrt(b)#

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

Which means that

#c=2+sqrt(7) or c=2-sqrt(7)#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We can use this with #a=c-2# and #b=sqrt(7)# as follows.

Given:

#(c-3)(c-1) = 6#

Multiply out the left hand side to get:

#c^2-4c+3 = 6#

Subtract #6# from both sides to get:

#c^2-4c-3 = 0#

Complete the square and use the difference of squares identity to find:

#0 = c^2-4c-3#

#color(white)(0) = c^2-4c+4-7#

#color(white)(0) = (c-2)^2-(sqrt(7))^2#

#color(white)(0) = ((c-2)-sqrt(7))((c-2)+sqrt(7))#

#color(white)(0) = (c-2-sqrt(7))(c-2+sqrt(7))#

Hence:

#c = 2+sqrt(7)" "# or #" "c = 2 - sqrt(7)#