Question 108379
{{{-(1/2)x^2+4x-6=0}}}
Let's multiply both sides by (-2) to get rid of the fraction. 
{{{(-2)(-(1/2)x^2+4x-6)=0}}}
{{{x^2-8x+12=0}}}Simplify.
We want to find a constant, a, such that,
{{{(x+a)^2=x^2+2ax+a^2}}}Expand the perfect square equation. 
Compare coefficients of that equation with your equation, 
{{{2ax=-8x}}}
{{{a=-4}}}
If a=-4, the {{{a^2=16}}}. 
In order get get 16 on the left hand side of the equation, that is, in order to complete the square, you need to add 4 to the left hand side. 
Whatever you do to one side of the equation, you must do to the other side. 
{{{x^2-8x+12=0}}}
{{{x^2-8x+16=4}}}Add 4 to both sides.
{{{(x-4)^2=4}}}Complete the square. 
{{{x-4=2}}} and {{{x-4=-2}}}
{{{x-4=2}}}
{{{highlight(x=6)}}}
{{{x-4=-2}}}
{{{highlight(x=2)}}}
Check your answers. 
{{{-(1/2)x^2+4x-6=0}}}
{{{-(1/2)(6)^2+4(6)-6=0}}}
{{{-18+24-6=0}}}
{{{0=0}}}
{{{-(1/2)x^2+4x-6=0}}}
{{{-(1/2)(2)^2+4(2)-6=0}}}
{{{-2+8-6=0}}}
{{{0=0}}}
Both answers lead to true statements.
Good answers.