Calculate $\sqrt{2x+5}$ to the power of $2$ and get $2x+5$.
$$2x+5=\left(-1-x\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(-1-x\right)^{2}$.
$$2x+5=1+2x+x^{2}$$
Subtract $1$ from both sides.
$$2x+5-1=2x+x^{2}$$
Subtract $1$ from $5$ to get $4$.
$$2x+4=2x+x^{2}$$
Subtract $2x$ from both sides.
$$2x+4-2x=x^{2}$$
Combine $2x$ and $-2x$ to get $0$.
$$4=x^{2}$$
Swap sides so that all variable terms are on the left hand side.
$$x^{2}=4$$
Subtract $4$ from both sides.
$$x^{2}-4=0$$
Consider $x^{2}-4$. Rewrite $x^{2}-4$ as $x^{2}-2^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(x-2\right)\left(x+2\right)=0$$
To find equation solutions, solve $x-2=0$ and $x+2=0$.
$$x=2$$ $$x=-2$$
Substitute $2$ for $x$ in the equation $1+x+\sqrt{2x+5}=0$.
$$1+2+\sqrt{2\times 2+5}=0$$
Simplify. The value $x=2$ does not satisfy the equation.
$$6=0$$
Substitute $-2$ for $x$ in the equation $1+x+\sqrt{2x+5}=0$.
$$1-2+\sqrt{2\left(-2\right)+5}=0$$
Simplify. The value $x=-2$ satisfies the equation.
$$0=0$$
Equation $\sqrt{2x+5}=-x-1$ has a unique solution.
