Solve x^2=2x+5-5xz

$${ x }^{ 2 } =2x+5-5xz$$

Solve for z

$$z=-\frac{x}{5}+\frac{2}{5}+\frac{1}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$2x+5-5xz=x^{2}$$

Subtract 2x from both sides.

$$5-5xz=x^{2}-2x$$

Subtract 5 from both sides.

$$-5xz=x^{2}-2x-5$$

The equation is in standard form.

$$\left(-5x\right)z=x^{2}-2x-5$$

Divide both sides by -5x.

$$\frac{\left(-5x\right)z}{-5x}=\frac{x^{2}-2x-5}{-5x}$$

Dividing by -5x undoes the multiplication by -5x.

$$z=\frac{x^{2}-2x-5}{-5x}$$

Divide x^{2}-2x-5 by -5x.

$$z=-\frac{x}{5}+\frac{2}{5}+\frac{1}{x}$$

Solve for x

$$x=\frac{\sqrt{25z^{2}-20z+24}}{2}-\frac{5z}{2}+1$$

$$x=-\frac{\sqrt{25z^{2}-20z+24}}{2}-\frac{5z}{2}+1$$