Question 577118


{{{x^2-10x+25=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-10}}}, and {{{c=25}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-10) +- sqrt( (-10)^2-4(1)(25) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-10}}}, and {{{c=25}}}



{{{x = (10 +- sqrt( (-10)^2-4(1)(25) ))/(2(1))}}} Negate {{{-10}}} to get {{{10}}}. 



{{{x = (10 +- sqrt( 100-4(1)(25) ))/(2(1))}}} Square {{{-10}}} to get {{{100}}}. 



{{{x = (10 +- sqrt( 100-100 ))/(2(1))}}} Multiply {{{4(1)(25)}}} to get {{{100}}}



{{{x = (10 +- sqrt( 0 ))/(2(1))}}} Subtract {{{100}}} from {{{100}}} to get {{{0}}}



{{{x = (10 +- sqrt( 0 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (10 +- 0)/(2)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{x = (10 + 0)/(2)}}} or {{{x = (10 - 0)/(2)}}} Break up the expression. 



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



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



So the only answer is {{{x = 5}}}