To solve this quadratic by completing the square, you first need to get your quadratic to the form
#color(blue)(x^2 + b/a = -c/a)#
In your case, this means adding #2# to both sides of the equation and dividing everything by #5#
#5x^2 + 8x - color(red)cancelcolor(black)(2) + color(red)cancelcolor(black)(2) = 0 + 2#
#(color(red)cancelcolor(black)(5)x^2)/color(red)cancelcolor(black)(5) + 8/5x = 2/5#
#x^2 + 8/5x = 2/5#
Next, use the coefficient of the #x#-term to help you determine what term needs to be added to both sides of the equation so that the left side can be written as the square of a binomial.
More specifically, divide said coefficient by #2#, square the result, then add it to both sides of the equation
#(8/5 * 1/2)^2 = 16/25#
Your equation will now look like this
#x^2 + 8/5x + 16/25 = 2/5 + 16/25#
The left side of the equation can be written as
#x^2 + 2 * (4/5) * x + (4/5)^2 = (x + 4/5)^2#
This means that you now have
#(x + 4/5)^2 = 26/25#
Take the square root of both sides to get
#sqrt( (x+4/5)^2) = sqrt(26/25)#
#x + 4/5 = +- sqrt(26)/5 implies x_(1,2) = -4/5 +- sqrt(26)/5#
The two solutions to your quadratic will be
#x_1 = color(green)(-4/5 - sqrt(26)/5)# and #x_2 = color(green)(-4/5 + sqrt(26)/5)#
