Question 1039626
Multiply both sides by  {{{(2x+2)(3x-2)}}}, the simplest or lowest common denominator.  This gives   {{{(x+2)(3x-2)=(3x+2)(2x+2)}}}


{{{3x^2+6x-2x-4=6x^2+2x+6x+4}}}


{{{4x-4=3x^2+8x+4}}}


{{{0=3x^2+4x+8}}}, but the discriminant is  {{{4^2-4*3*8=-80}}}, a negative numbers.  NO REAL SOLUTIONS.  Do you want Complex solutions?



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"Yes"...


You should both check the discussion for this in YOUR textbook, and also review this couple of lessons:
<a href="https://www.algebra.com/my/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev">Solve Quadratic Equation using Completing The Square</a>
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<a href="https://www.algebra.com/my/Completing-the-Square-Using-Pictures.lesson?content_action=show_dev">What is Completing The Square</a>


Going directly to the general solution formula for a quadratic equation,
{{{x=(-8+- sqrt(-80))/(2*3)}}}


{{{x=(-8+- sqrt(-1*2*2*2*2*5))/(2*3)}}}


{{{x=(-8+- 4i*sqrt(5))/(2*3)}}}


{{{highlight(x=(-4+- 2i*sqrt(5))/3)}}}
Those will be no trouble for the denominator because the solutions will not make denominators become 0.