Question 254013


{{{x^2-6x+7=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-6x+7}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-6}}}, and {{{C=7}}}



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



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



{{{x = (-(-6) +- sqrt( (-6)^2-4(1)(7) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-6}}}, and {{{C=7}}}



{{{x = (6 +- sqrt( (-6)^2-4(1)(7) ))/(2(1))}}} Negate {{{-6}}} to get {{{6}}}. 



{{{x = (6 +- sqrt( 36-4(1)(7) ))/(2(1))}}} Square {{{-6}}} to get {{{36}}}. 



{{{x = (6 +- sqrt( 36-28 ))/(2(1))}}} Multiply {{{4(1)(7)}}} to get {{{28}}}



{{{x = (6 +- sqrt( 8 ))/(2(1))}}} Subtract {{{28}}} from {{{36}}} to get {{{8}}}



{{{x = (6 +- sqrt( 8 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (6 +- 2*sqrt(2))/(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 = (6)/(2) +- (2*sqrt(2))/(2)}}} Break up the fraction.  



{{{x = 3 +- sqrt(2)}}} Reduce.  



{{{x = 3+sqrt(2)}}} or {{{x = 3-sqrt(2)}}} Break up the expression.  



So the solutions are {{{x = 3+sqrt(2)}}} or {{{x = 3-sqrt(2)}}} 



which approximate to {{{x=4.414}}} or {{{x=1.586}}}