Question 200758

{{{x^2-36=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-36}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=0}}}, and {{{C=-36}}}



Let's use the quadratic formula to solve for "x":



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



{{{x = (0 +- sqrt( (0)^2-4(1)(-36) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=0}}}, and {{{C=-36}}}



{{{x = (0 +- sqrt( 0-4(1)(-36) ))/(2(1))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0--144 ))/(2(1))}}} Multiply {{{4(1)(-36)}}} to get {{{-144}}}



{{{x = (0 +- sqrt( 0+144 ))/(2(1))}}} Rewrite {{{sqrt(0--144)}}} as {{{sqrt(0+144)}}}



{{{x = (0 +- sqrt( 144 ))/(2(1))}}} Add {{{0}}} to {{{144}}} to get {{{144}}}



{{{x = (0 +- sqrt( 144 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (0 +- 12)/(2)}}} Take the square root of {{{144}}} to get {{{12}}}. 



{{{x = (0 + 12)/(2)}}} or {{{x = (0 - 12)/(2)}}} Break up the expression. 



{{{x = (12)/(2)}}} or {{{x =  (-12)/(2)}}} Combine like terms. 



{{{x = 6}}} or {{{x = -6}}} Simplify. 



So the solutions are {{{x = 6}}} or {{{x = -6}}}