Question 164558
The best way to find the roots of a quadratic equation (values that make the ENTIRE equation equal to zero) is to use the quadratic formula.



{{{x^2-6x+8=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-6}}}, and {{{c=8}}}



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)(8) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-6}}}, and {{{c=8}}}



{{{x = (6 +- sqrt( (-6)^2-4(1)(8) ))/(2(1))}}} Negate {{{-6}}} to get {{{6}}}. 



{{{x = (6 +- sqrt( 36-4(1)(8) ))/(2(1))}}} Square {{{-6}}} to get {{{36}}}. 



{{{x = (6 +- sqrt( 36-32 ))/(2(1))}}} Multiply {{{4(1)(8)}}} to get {{{32}}}



{{{x = (6 +- sqrt( 4 ))/(2(1))}}} Subtract {{{32}}} from {{{36}}} to get {{{4}}}



{{{x = (6 +- sqrt( 4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



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



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



{{{x = (8)/(2)}}} or {{{x =  (4)/(2)}}} Combine like terms. 



{{{x = 4}}} or {{{x = 2}}} Simplify. 



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Answer:


So the solutions are {{{x = 4}}} or {{{x = 2}}}