First, expand the term on the left using this special rule for multiplying quadratics:

#(color(red)(x) + color(blue)(y))^2 = color(red)(x)^2 + 2color(red)(x)color(blue)(y) + color(blue)(y)^2#

Substituting gives:

#(color(red)(-2x) + color(blue)(5))^2 = -8#

#(color(red)(-2x))^2 + (2 * color(red)(-2x) * color(blue)(5)) + color(blue)(5)^2 = -8#

#4x^2 + (-20x) + 25 = -8#

#4x^2 - 20x + 25 = -8#

We can next convert this to standard form:

#4x^2 - 20x + 25 + color(red)(8) = -8 + color(red)(8)#

#4x^2 - 20x + 33 = 0#

We can now use the quadratic formula to find the solutions for #x#. The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(4)# for #color(red)(a)#

#color(blue)(-20)# for #color(blue)(b)#

#color(green)(33)# for #color(green)(c)# gives:

#x = (-(color(blue)(-20)) +- sqrt(color(blue)(-20)^2 - (4 * color(red)(4) * color(green)(33))))/(2 * color(red)(4))#

#x = (color(blue)(20) +- sqrt(400 - 528))/8#

#x = (color(blue)(20) +- sqrt(-128))/8#

#x = (color(blue)(20) +- sqrt(64 * -2))/8#

#x = (color(blue)(20) +- sqrt(64)sqrt(-2))/8#

#x = (color(blue)(20) +- 8sqrt(-2))/8#

Or

#x = color(blue)(20)/8 +- (8sqrt(-2))/8#

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