Question 325471
You're off to a great start. You just need to solve {{{x^2 - 10x +2=0}}} now.



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



Notice that the quadratic {{{x^2-10x+2}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-10}}}, and {{{C=2}}}



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)(2) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-10}}}, and {{{C=2}}}



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



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



{{{x = (10 +- sqrt( 100-8 ))/(2(1))}}} Multiply {{{4(1)(2)}}} to get {{{8}}}



{{{x = (10 +- sqrt( 92 ))/(2(1))}}} Subtract {{{8}}} from {{{100}}} to get {{{92}}}



{{{x = (10 +- sqrt( 92 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (10 +- 2*sqrt(23))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (10)/(2) +- (2*sqrt(23))/(2)}}} Break up the fraction.  



{{{x = 5 +- sqrt(23)}}} Reduce.  



{{{x = 5+sqrt(23)}}} or {{{x = 5-sqrt(23)}}} Break up the expression.  



So the solutions are {{{x = 5+sqrt(23)}}} or {{{x = 5-sqrt(23)}}}