Question 200196


{{{x^2+12x-64=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+12x-64}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=12}}}, and {{{C=-64}}}



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



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



{{{x = (-(12) +- sqrt( (12)^2-4(1)(-64) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=12}}}, and {{{C=-64}}}



{{{x = (-12 +- sqrt( 144-4(1)(-64) ))/(2(1))}}} Square {{{12}}} to get {{{144}}}. 



{{{x = (-12 +- sqrt( 144--256 ))/(2(1))}}} Multiply {{{4(1)(-64)}}} to get {{{-256}}}



{{{x = (-12 +- sqrt( 144+256 ))/(2(1))}}} Rewrite {{{sqrt(144--256)}}} as {{{sqrt(144+256)}}}



{{{x = (-12 +- sqrt( 400 ))/(2(1))}}} Add {{{144}}} to {{{256}}} to get {{{400}}}



{{{x = (-12 +- sqrt( 400 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-12 +- 20)/(2)}}} Take the square root of {{{400}}} to get {{{20}}}. 



{{{x = (-12 + 20)/(2)}}} or {{{x = (-12 - 20)/(2)}}} Break up the expression. 



{{{x = (8)/(2)}}} or {{{x =  (-32)/(2)}}} Combine like terms. 



{{{x = 4}}} or {{{x = -16}}} Simplify. 



So the solutions are {{{x = 4}}} or {{{x = -16}}}