Question 263641


{{{x^2-8x+3=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-8x+3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-8}}}, and {{{C=3}}}



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)(3) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-8}}}, and {{{C=3}}}



{{{x = (8 +- sqrt( (-8)^2-4(1)(3) ))/(2(1))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{x = (8 +- sqrt( 64-4(1)(3) ))/(2(1))}}} Square {{{-8}}} to get {{{64}}}. 



{{{x = (8 +- sqrt( 64-12 ))/(2(1))}}} Multiply {{{4(1)(3)}}} to get {{{12}}}



{{{x = (8 +- sqrt( 52 ))/(2(1))}}} Subtract {{{12}}} from {{{64}}} to get {{{52}}}



{{{x = (8 +- sqrt( 52 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (8 +- 2*sqrt(13))/(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 = (8)/(2) +- (2*sqrt(13))/(2)}}} Break up the fraction.  



{{{x = 4 +- sqrt(13)}}} Reduce.  



{{{x = 4+sqrt(13)}}} or {{{x = 4-sqrt(13)}}} Break up the expression.  



So the solutions are {{{x = 4+sqrt(13)}}} or {{{x = 4-sqrt(13)}}}