#### Answer

$x=1$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{5-x}=x+1
,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation.
$\bf{\text{Solution Details:}}$
Squaring both sides of the equation results to \begin{array}{l}\require{cancel}
\left( \sqrt{5-x} \right)^2=(x+1)^2
\\\\
5-x=(x+1)^2
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel}
5-x=(x)^2+2(x)(1)+(1)^2
\\\\
5-x=x^2+2x+1
.\end{array}
Using the propertis of equality, in the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
-x^2+(-x-2x)+(5-1)=0
\\\\
-x^2-3x+4=0
\\\\
-1(-x^2-3x+4)=(0)(-1)
\\\\
x^2+3x-4=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is
\begin{array}{l}\require{cancel}
(x+4)(x-1)=0
.\end{array}
Equating each factor to zero (Zero Product Property) results to
\begin{array}{l}\require{cancel}
x+4=0
\\\\\text{OR}\\\\
x-1=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x+4=0
\\\\
x=-4
\\\\\text{OR}\\\\
x-1=0
\\\\
x=1
.\end{array}
Upon checking, only $
x=1
$ satisfies the original equation.