Square both sides of the equation to remove the radical.
$x + 7 = x^2 + 2x + 1$

Set the equation equal to $0 = x^2 + x − 6$$0$ because it is quadratic (there is an $x^2$-term.)

Use the Zero Product Property to solve.

Check your solutions in the original equation. Do they both work?

Notice that you can factor $x^2 + 3x$ to $x(x + 3)$.

Multiply both sides of the equation by a common denominator to remove the fractions.
$x(x+3)\left (\frac{2}{x+3}-\frac{1}{x} \right )\ =\ x(x+3)\left (\frac{-6}{x\left (x+3 \right )} \right )$
$2x - (x + 3) = -6$

Solve for $x$.
$x = -3$

Substitute your solution back into the original equation. Is $x = -3$ a valid solution to this problem?