Question 603119
You can solve {{{x^2-6x+4=0}}} by "completing the square" or you can apply the quadratic formula.
 
COMPLETING THE SQUARE:
{{{x^2-6x+4=0}}} --> {{{x^2-6x=-4}}}
At this point you would recognize that the left hand side of the equation resembles the square
{{{(x-3)^2=x^2-6x+9}}}
So, you add 9 to both sides of {{{x^2-6x=-4}}} to get the equivalent equation with the complete square on the left hand side:
{{{x^2-6x+9=-4+9}}} --> {{{x^2-6x+9=5}}} ---> {{{(x-3)^2=5}}}
At this point, you know that either
{{{x-3=sqrt(5)}}} or {{{x-3=-sqrt(5)}}}
Adding 3 to both sides of the two equations above, you get the two solutions
{{{x=3+sqrt(5)}}} and {{{x=3-sqrt(5)}}},
which can be summarized as
{{{x=3 +- sqrt(5)}}}
 
THE QUADRATIC FORMULA
All quadratic equations are of the form
{{{ax^2+bx+c=0}}} where a, b, and c are the coefficients which could be any number, with the only restriction that a cannot be zero.
The solution is given by the quadratic formula:
{{{x=(-b +- sqrt(b^2-4*a*c))/(2*a)}}}
In the equation {{{x^2-6x+4=0}}},
a=1, b=-6 and c=4, so the quadratic formula gives us
{{{x=(-(-6) +- sqrt((-6)^2-4*1*4))/(2*1)=(6 +- sqrt(36-16))/2=(6 +- sqrt(20))/2}}}
That can be simplified like this
{{{x=(6 +- sqrt(20))/2=(6 +- sqrt(4*5))/2=(6 +- sqrt(4)*sqrt(5))/2=(6 +- 2*sqrt(5))/2=2(3 +- sqrt(5))/2=3 +- sqrt(5)}}}