Question 196637
{{{x^2= 8x-16 }}} Start with the given equation.



{{{x^2-8x=-16 }}} Subtract 8x from both sides.



{{{x^2-8x+16=0 }}} Add 16 to both sides.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-8}}}, and {{{C=16}}}



Let's use the quadratic formula to solve for x



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-8) +- sqrt( (-8)^2-4(1)(16) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-8}}}, and {{{C=16}}}



{{{x = (8 +- sqrt( (-8)^2-4(1)(16) ))/(2(1))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{x = (8 +- sqrt( 64-4(1)(16) ))/(2(1))}}} Square {{{-8}}} to get {{{64}}}. 



{{{x = (8 +- sqrt( 64-64 ))/(2(1))}}} Multiply {{{4(1)(16)}}} to get {{{64}}}



{{{x = (8 +- sqrt( 0 ))/(2(1))}}} Subtract {{{64}}} from {{{64}}} to get {{{0}}}



{{{x = (8 +- sqrt( 0 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (8 +- 0)/(2)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{x = (8 + 0)/(2)}}} or {{{x = (8 - 0)/(2)}}} Break up the expression. 



{{{x = (8)/(2)}}} or {{{x =  (8)/(2)}}} Combine like terms. 



{{{x = 4}}} or {{{x = 4}}} Simplify. 



So the answer is {{{x = 4}}} (with a multiplicity of 2)