Isolate the radical term, then square both sides:
[tex]2x+\sqrt{x+1}=8[/tex]
[tex]\sqrt{x+1}=8-2x[/tex]
Note that [tex]\sqrt{x+1}[/tex] is non-negative and requires that [tex]x\ge-1[/tex] in order to be defined, while [tex]8-2x<0[/tex] when [tex]x>4[/tex]. This means we can only find real solutions in the range [tex]-1\le x\le4[/tex].
[tex](\sqrt{x+1})^2=(8-2x)^2[/tex]
[tex]x+1=(8-2x)^2[/tex]
Now just expand the RHS and simplify as much as possible:
[tex]x+1=64-32x+4x^2[/tex]
[tex]4x^2-33x+63=0[/tex]
[tex](x-3)(4x-21)=0[/tex]
[tex]\implies x=3,x=\dfrac{21}4[/tex]
Now, [tex]\dfrac{21}4=5.25[/tex], which is larger than 4, so we omit that solution. Then [tex]x=3[/tex] is the only solution.
