Question 62205
Solve by completing the square:
{{{x^2-5x+2 = 0}}}?  You left out the = so I'm presuming it was supposed to be = 0. Start by subtracting 2 from both sides.
{{{x^2-5x = -2}}} Now complete the square in the x-terms by adding the square of half the x-coefficient, that's {{{(5/2)^2 = 25/4}}} to both sides.
{{{x^2-5x+25/4 = (25/4)-2}}} Factor the left sides and simplify.
{{{(x-5/2)^2 = (25-8)/4}}}
{{{(x-5/2)^2 = 17/4}}}  Take the square root of both sides.
{{{x-5/2 = (sqrt(17))/2}}} (+ or -)  Finally, add 5/2 to both sides.
{{{x = 5/2+(1/2)sqrt(17)}}} and
{{{x = 5/2-(1/2)sqrt(17)}}}