Question 152·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

The principal square root is never negative, so the right side of the given equation must satisfy $m-3\ge0$. That gives $m\ge3$. Squaring can lose this sign condition.

Replace each $m$ with $x_1$ and type $\sqrt{x_1+9}\sim x_1-3\{x_1\ge3\}$. The $\sim$ tells Desmos to solve for $x_1$; the braces keep the allowed values. Under PARAMETERS, Desmos shows $x_1=7$. You entered the original equation, so $m=7$ is a solution. Grid in 7.

Square both entire sides of the equation to remove the square root: $m+9=(m-3)^2$. Squaring can make opposite numbers look equal, so the results will need checking.

Use $(a-b)^2=a^2-2ab+b^2$. The middle term here is $-2(m)(3)=-6m$. Expand:

Subtract $m+9$ from both sides so you can factor the resulting equation:

Both terms contain $m$, so take it out as a common factor:

The zero-product property says that if a product is $0$, at least one factor is $0$. Set each factor equal to $0$: $m=0\quad\text{or}\quad m-7=0$. Add $7$ to the second equation: $m=0\quad\text{or}\quad m=7$. These are candidates, not yet confirmed solutions.

Type each candidate into both sides of the original equation. Desmos shows $3$ and $-3$ when $m=0$, so that value fails. It shows $4$ and $4$ when $m=7$, so that value works. The solution is $m=7$. Grid in 7.