Question 153512
First we need to change the equation into the standard form.

{{{(1/3)x^2-3x+14+2/3=2x}}}
{{{(1/3)x^2-3x+44/3=2x}}}
{{{3((1/3)x^2-3x+44/3)=3*2x}}}    (multiply both sides by 3)
{{{x^2-9x+44=6x}}}
{{{x^2-15x+44=0}}}
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SOLVED BY FACTORING
The left side of the equation can be factored as {{{x^2-15x+44=(x-4)(x-11)}}}.
So the equation becomes:
{{{(x-4)(x-11)=0}}}
so x= 4 or x = 11
SOLVED BY USING QUADRATIC FORMULA
Substituting 
a = 1
b = -15
c = 44
into
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}, we have
{{{x = (-(-15) +- sqrt( (-15)^2-4*1*44 ))/(2*1) }}}
{{{x = (15 +- sqrt( 225-176 ))/2 }}}
{{{x = (15 +- sqrt( 49 ))/2 }}}
{{{x = (15 +- 7)/2 }}}
So
{{{x = (15 + 7)/2 =22/2 = 11}}}
Or
{{{x = (15 - 7)/2 =8/2 = 4}}}