Well, a "tricky" solution could involve taking the negative square root of (x+13), so that's included here, though sqrt is presumed to be the positive square root.

syms x

fplot(@(x)x^2-13)

hold on

fplot(sqrt(x+13))

fplot(-sqrt(x+13))

hold off

grid

eq1=x^2-13==sqrt(x+13);

eq2=x^2-13==-sqrt(x+13);

s=solve(eq1 ,x)

eval(s)

s2=solve(eq2 ,x)

eval(s2)

When no tricks are involved, x=-4 or x=sqrt(53)/2 + 1/2 =~4.14005494464026.

Getting "tricky" I assume involves allowing x=3 or x=1/2 - sqrt(53)/2=~-3.14005494464026. This is nonstandard.