Solve x-q==z^2+(x^2+y^2)

$$x - q = = z ^ { 2 } + ( x ^ { 2 } + y ^ { 2 } )$$

Solve for q

$$q=x-z^{2}-y^{2}-x^{2}$$

Steps for Solving Linear Equation

Subtract x from both sides.

$$-q=z^{2}+x^{2}+y^{2}-x$$

The equation is in standard form.

$$-q=x^{2}-x+y^{2}+z^{2}$$

Divide both sides by -1.

$$\frac{-q}{-1}=\frac{x^{2}-x+y^{2}+z^{2}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$q=\frac{x^{2}-x+y^{2}+z^{2}}{-1}$$

Divide -x+z^{2}+x^{2}+y^{2} by -1.

$$q=x-z^{2}-y^{2}-x^{2}$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{1-4q-4z^{2}-4y^{2}}+1}{2}$$

$$x=\frac{\sqrt{1-4q-4z^{2}-4y^{2}}+1}{2}$$

Solve for x

$$x=\frac{-\sqrt{1-4q-4z^{2}-4y^{2}}+1}{2}$$

$$x=\frac{\sqrt{1-4q-4z^{2}-4y^{2}}+1}{2}\text{, }q\leq \frac{1}{4}-z^{2}-y^{2}$$