Solve 3x-2(z(x))=6x^2+3x+4

$$3x-2(z(x)) = 6 { x }^{ 2 } +3x+4$$

Solve for z

$$z=-3x-\frac{2}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 3x from both sides.

$$-2zx=6x^{2}+3x+4-3x$$

Combine 3x and -3x to get 0.

$$-2zx=6x^{2}+4$$

The equation is in standard form.

$$\left(-2x\right)z=6x^{2}+4$$

Divide both sides by -2x.

$$\frac{\left(-2x\right)z}{-2x}=\frac{6x^{2}+4}{-2x}$$

Dividing by -2x undoes the multiplication by -2x.

$$z=\frac{6x^{2}+4}{-2x}$$

Divide 6x^{2}+4 by -2x.

$$z=-3x-\frac{2}{x}$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{z^{2}-24}-z}{6}$$

$$x=\frac{\sqrt{z^{2}-24}-z}{6}$$

Solve for x

$$x=\frac{-\sqrt{z^{2}-24}-z}{6}$$

$$x=\frac{\sqrt{z^{2}-24}-z}{6}\text{, }|z|\geq 2\sqrt{6}$$