First of all, please be aware that #x != 0# and # x != -3# must hold so that the equation is defined.

Now, with these restrictions in mind, let's start solving. :)

First of all, we need to get rid of the denominators. To do so, multiply both sides with # x * (x+3)#:

#color(white)(xxxss) 5/x - 2 = 2 / (x+3)#

#<=> color(white)(xx) 5 ( x + 3 ) - 2 x (x + 3) = 2 x#

#<=> color(white)(xx) 5x + 15 - 2x^2 - 6x = 2x #

... add #-2x# on both sides and simplify ...

#<=> color(white)(xx) 15 - 3x - 2x^2 = 0#

This is a quadratic equation which can be solved for example with the quadratic formula which states that the solution of #ax^2 + bx + c = 0# is

# x = (-b +- sqrt(b^2-4ac))/(2a)#

In our case, #a = -2#, #b = -3# and #c = 15#.

This leads us to

#x = (3 +- sqrt(9 - 4 * (-2)*15))/-4 = (3 +- sqrt(129))/-4#

The two solutions are

#x = -3/4 + sqrt(129)/4 color(white)(xx)# and #color(white)(xx)x = -3/4 - sqrt(129)/4#