How do you use the quadratic formula to solve #2/3x^2-2=1/2x+1#?

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Answer 1 · 2017-07-25T13:00:58

#x =1.921#

#x = -1.171#

#2/3x^2-2=1/2x+1" "larr# get rid of the fractions: #xx6#

#(cancel6^2xx2)/cancel3x^2-6xx2=(cancel6^3xx1)/cancel2x+6xx1#

#4x^2-12=3x+6" "larr# make it equal to #0#

#4x^2-3x-18 =0" "larr# does not factorise

Use the quadratic formula to solve for #x#

#x =(-b+-sqrt(b^2-4ac))/(2a)" "# with #a=4, b=-3 and c=-18#

#x =(-(-3)+-sqrt((-3)^2-4(4)(-18)))/(2(4))#

#x =(3+-sqrt((9+144)))/(8)#

#x =(3+-sqrt153)/8#

Solve twice to find #x#

#x =1.921#

#x = -1.171#