Assuming you are trying to solve for #x#:

First, subtract #color(red)(a)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-color(red)(a) + a - 2x^2 = -color(red)(a) - 5#

#0 - 2x^2 = -a - 5#

#-2x^2 = -a - 5#

Next, divide each side of the equation by #color(red)(-2)# to isolate #x^2# while keeping the equation balanced:

#(-2x^2)/color(red)(-2) = (-a - 5)/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))x^2)/cancel(color(red)(-2)) = (-a)/-2 + (-5)/-2#

#x^2 = a/2 + 5/2#

#x^2 = (a + 5)/2#

Now, take the square root of each side of the equation to solve for #x# while keeping the equation balanced. Remember the square root of a number produces a positive and negative result:

#sqrt(x^2) = +-sqrt((a + 5)/2)#

#x = +-sqrt((a + 5)/2)#