Solve sqrt{m+1}+(n-3)^2=0

$$\sqrt { m + 1 } + ( n - 3 ) ^ { 2 } = 0$$

Solve for m

$$m=-1$$

$$n=3$$

Steps for Solving Radical Equations

Subtract \left(n-3\right)^{2} from both sides of the equation.

$$\sqrt{m+1}+\left(n-3\right)^{2}-\left(n-3\right)^{2}=-\left(n-3\right)^{2}$$

Subtracting \left(n-3\right)^{2} from itself leaves 0.

$$\sqrt{m+1}=-\left(n-3\right)^{2}$$

Square both sides of the equation.

$$m+1=\left(n-3\right)^{4}$$

Subtract 1 from both sides of the equation.

$$m+1-1=\left(n-3\right)^{4}-1$$

Subtracting 1 from itself leaves 0.

$$m=\left(n-3\right)^{4}-1$$

Subtract 1 from \left(n-3\right)^{4}.

$$m=\left(n-4\right)\left(n-2\right)\left(\left(n-3\right)^{2}+1\right)$$

Solve for m (complex solution)

$$m=-\left(4-n\right)\left(n-2\right)\left(\left(n-3\right)^{2}+1\right)$$

$$arg(\left(n-3\right)^{2})\geq \pi \text{ or }n=3$$

Solve for n (complex solution)

$$n=-i\sqrt[4]{m+1}+3$$

$$n=i\sqrt[4]{m+1}+3$$

Solve for n

$$n=3$$

$$m=-1$$