Question 690016
Your equation is in the correct form for factoring
{{{ (3/4)*x^2 - 3x + 1/2 = 0 }}}
Multiply both sides by {{{ 4 }}}
{{{ 3x^2 - 12x + 2 = 0 }}}
This is in the form
{{{ a*x^2 + b*x + c = 0 }}}
{{{ a = 3 }}}
{{{ b = -12 }}}
{{{ c = 2 }}}
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Quadratic formula:
{{{ x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ x = (-(-12) +- sqrt( (-12)^2-4*3*2 ))/(2*3) }}}
{{{ x = ( 12 +- sqrt( 144 - 24 )) / 6 }}}
{{{ x = ( 12 +- sqrt( 120 )) / 6 }}}
{{{ x =  ( 12 +- 10.95445 ) / 6 }}}
{{{ x =  ( 12 + 10.95445 ) / 6 }}}
{{{ x = 22.95445 / 6 }}}
(1) {{{ x = 3.82574 }}}
and, taking the negative square root,
{{{ x =  ( 12 - 10.95445 ) / 6 }}}
{{{ x = 1.04555 / 6 }}}
(2) {{{ x = .17426 }}}
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Rewrite (1) and (2) as
{{{ x - 3.82574 = 0 }}}
and
(2) {{{ x - .17426 = 0 }}}
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Now I can say:
{{{ ( x - 3.82574 )*( x - .17426 ) = 0 }}}
{{{ x^2 - 3.82574x - .17426x + ( -3.82574 )*( -.17426 ) = 0 }}}
{{{ x^2 - 3.99999x + .66667 = 0 }}}
If I multiply both sides by {{{ 3 }}}, I get 
what I started with,
{{{ 3x^2 - 12x + 2 = 0 }}}
Now if I divide through by {{{ 4 }}}, 
I get your original equation,
{{{ (3/4)*x^2 - 3x + 1/2 = 0 }}}
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A general explanation would be:
What you have is:
( a parabola function ) = 0
The function is only zero where it crosses
the x-axis. You cannot have more than
2 real solutions with a parabola.
( you might have 1, or none )
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In the quadratic formula,  {{{ b^2-4*a*c }}}
is the discriminant
It tells you the nature of the roots, so you have:
x[1] = -b/(2a) + (sqrt( discriminant) ) / (2a)
x[2] = -b/(2a) - (sqrt( discriminant) ) / (2a)
x[1] and x[2] are the roots, or x-crossings
The 1st term, -b/(2a), is the x-coordinate of the 
vertex.
Now you see that you have
x[1] = ( x of the vertex ) + ( a displacement to the right of the vertex )
x[2] = ( x of the vertex ) - ( a displacement to the left of the vertex )
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If you ended up with solutions {{{ x = 9 }}}
and {{{ x = 5 }}}, then you can see that {{{ x = 7 }}}
is the mid-point and that is the x-coordinate of the vertex.
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Hope this helps