Question 397042


{{{x^2-4x-5=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-4x-5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-4}}}, and {{{C=-5}}}



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



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



{{{x = (-(-4) +- sqrt( (-4)^2-4(1)(-5) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-4}}}, and {{{C=-5}}}



{{{x = (4 +- sqrt( (-4)^2-4(1)(-5) ))/(2(1))}}} Negate {{{-4}}} to get {{{4}}}. 



{{{x = (4 +- sqrt( 16-4(1)(-5) ))/(2(1))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16--20 ))/(2(1))}}} Multiply {{{4(1)(-5)}}} to get {{{-20}}}



{{{x = (4 +- sqrt( 16+20 ))/(2(1))}}} Rewrite {{{sqrt(16--20)}}} as {{{sqrt(16+20)}}}



{{{x = (4 +- sqrt( 36 ))/(2(1))}}} Add {{{16}}} to {{{20}}} to get {{{36}}}



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



{{{x = (4 +- 6)/(2)}}} Take the square root of {{{36}}} to get {{{6}}}. 



{{{x = (4 + 6)/(2)}}} or {{{x = (4 - 6)/(2)}}} Break up the expression. 



{{{x = (10)/(2)}}} or {{{x =  (-2)/(2)}}} Combine like terms. 



{{{x = 5}}} or {{{x = -1}}} Simplify. 



So the solutions are {{{x = 5}}} or {{{x = -1}}} 

  

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