How do you evaluate #\frac { 1} { 2} x ^ { 2} + \frac { 2} { 3} = \frac { 1} { 3} x - \frac { 3} { 2}#?

1 Answer
Pankaj Solanki
Sep 16, 2017

#x=(1+-isqrt(38))/(3)#

Explanation:

#x^2/2+2/3=x/3-3/2#
Bring all terms on one side
#x^2/2+2/3-x/3+3/2=0#
Take L.C.M. 6
#(3x^2+4-2x+9)/6=0#
#3x^2+4-2x+9=0#
#3x^2-2x+13=0#
#x=(-(-2)+-sqrt((-2)^2-4(3)(13)))/(2(3))#
#x=(2+-sqrt(4-156))/(6)#
#x=(2+-sqrt(-152))/(6)#
#x=(1+-sqrt(-38))/(3)#
#x=(1+-isqrt(38))/(3)#

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