Question 538112
{{{x^2-8x-9=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-8x-9}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-8}}}, and {{{C=-9}}}



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)(-9) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-8}}}, and {{{C=-9}}}



{{{x = (8 +- sqrt( (-8)^2-4(1)(-9) ))/(2(1))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{x = (8 +- sqrt( 64-4(1)(-9) ))/(2(1))}}} Square {{{-8}}} to get {{{64}}}. 



{{{x = (8 +- sqrt( 64--36 ))/(2(1))}}} Multiply {{{4(1)(-9)}}} to get {{{-36}}}



{{{x = (8 +- sqrt( 64+36 ))/(2(1))}}} Rewrite {{{sqrt(64--36)}}} as {{{sqrt(64+36)}}}



{{{x = (8 +- sqrt( 100 ))/(2(1))}}} Add {{{64}}} to {{{36}}} to get {{{100}}}



{{{x = (8 +- sqrt( 100 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (8 +- 10)/(2)}}} Take the square root of {{{100}}} to get {{{10}}}. 



{{{x = (8 + 10)/(2)}}} or {{{x = (8 - 10)/(2)}}} Break up the expression. 



{{{x = (18)/(2)}}} or {{{x =  (-2)/(2)}}} Combine like terms. 



{{{x = 9}}} or {{{x = -1}}} Simplify. 



So the solutions are {{{x = 9}}} or {{{x = -1}}}



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