How do you solve #2/3 x^2 + 1/4 x= 3#?

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Answer 1 · 2016-01-25T12:23:24

#x=(-3+-3sqrt(129))/16#

First simplify #2/3x^2+1/4x=3# by multiplying everything by #12# to clear the fractions:

#8x^2+3x=36#

Re-write in standard #ax^2+bx+c=0# form
#8x^2+3x-36=0#

Apply the quadratic formula #x=(-b+-sqrt(b^2-4ac))/(2a)# to obtain the roots.

#x=(-3+-sqrt(9-4(8)(-36)))/(2(8))#

Simplify
#x=-(3+-3sqrt(129))/16#