First, we can square both sides of the equation to eliminate the fractional exponent while keeping the equation balanced:

#((5 - x)^(1/2))^2 = (x + 1)^2#

#((5 - x)^1 = x^2 + 2x + 1#

#5 - x = x^2 + 2x + 1#

Next, we can convert the equation into standard form:

#5 - x - color(red)(5) + color(blue)(x) = x^2 + 2x + 1 - color(red)(5) + color(blue)(x)#

#5 - color(red)(5) - x + color(blue)(x) = x^2 + 2x + color(blue)(x) + 1 - color(red)(5)#

#0 - 0 = x^2 + 3x - 4#

#0 = x^2 + 3x - 4#

#x^2 + 3x - 4 = 0#

Now, we can use the quadratic equation to solve this problem:

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)(1)# for #color(red)(a)#

#color(blue)(3)# for #color(blue)(b)#

#color(green)(-4)# for #color(green)(c)# gives:

#x = (-color(blue)(3) +- sqrt(color(blue)(3)^2 - (4 * color(red)(1) * color(green)(-4))))/(2 * color(red)(1))#

#x = (-color(blue)(3) +- sqrt(9 - (-16)))/2#

#x = (-color(blue)(3) - sqrt(9 - (-16)))/2# and #x = (-color(blue)(3) + sqrt(9 - (-16)))/2#

#x = (-color(blue)(3) - sqrt(9 + 16))/2# and #x = (-color(blue)(3) + sqrt(9 + 16))/2#

#x = (-color(blue)(3) - sqrt(25))/2# and #x = (-color(blue)(3) + sqrt(25))/2#

#x = (-color(blue)(3) - 5)/2# and #x = (-color(blue)(3) + 5)/2#

#x = -8/2# and #x = 2/2#

#x = -4# and #x = 1#

Substituting both solutions into the original equation gives:

#(5 - (-4))^(1/2) = -4 + 1# and #(5 - 1)^(1/2) = 1 + 1#

#(5 + 4)^(1/2) = -3# and #(4)^(1/2) = 2#

#(9)^(1/2) = -3# and #2 = 2#

#3 != -3# and #2 = 2#

Therefore the solution #x = -4# is extraneous.