Solve (3x-9)^2+sqrt{y+1}+|z-5|=0

$$( 3 x - 9 ) ^ { 2 } + \sqrt { y + 1 } + | z - 5 | = 0$$

Solve for y

$$\left\{\begin{matrix}y=-\left(-9\left(x-3\right)^{2}-z+6\right)\left(9\left(x-3\right)^{2}+z-4\right)\text{, }&-9\left(x-3\right)^{2}-z+5\geq 0\text{ and }z\geq 5\\y=-\left(-9\left(x-3\right)^{2}+z-4\right)\left(9\left(x-3\right)^{2}-z+6\right)\text{, }&-\left(9\left(x-3\right)^{2}-z+5\right)\geq 0\text{ and }z\leq 5\end{matrix}\right.$$

Steps for Solving Radical Equations

Subtract \left(3x-9\right)^{2}+|z-5| from both sides of the equation.

$$\sqrt{y+1}+\left(3x-9\right)^{2}+|z-5|-\left(\left(3x-9\right)^{2}+|z-5|\right)=-\left(\left(3x-9\right)^{2}+|z-5|\right)$$

Subtracting \left(3x-9\right)^{2}+|z-5| from itself leaves 0.

$$\sqrt{y+1}=-\left(\left(3x-9\right)^{2}+|z-5|\right)$$

Subtract \left(3x-9\right)^{2}+|z-5| from 0.

$$\sqrt{y+1}=-\left(9\left(x-3\right)^{2}+|z-5|\right)$$

Square both sides of the equation.

$$y+1=\left(9\left(x-3\right)^{2}+|z-5|\right)^{2}$$

Subtract 1 from both sides of the equation.

$$y+1-1=\left(9\left(x-3\right)^{2}+|z-5|\right)^{2}-1$$

Subtracting 1 from itself leaves 0.

$$y=\left(9\left(x-3\right)^{2}+|z-5|\right)^{2}-1$$

Solve for x

$$x=3$$

$$y=-1\text{ and }z=5$$