Solve left(1-pright)x^2+5=2x

$$\left( 1-p \right) { x }^{ 2 } +5 = 2x$$

Solve for p

$$p=\frac{x^{2}-2x+5}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 1-p by x^{2}.

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

Subtract x^{2} from both sides.

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

Subtract 5 from both sides.

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

The equation is in standard form.

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

Divide both sides by -x^{2}.

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

Dividing by -x^{2} undoes the multiplication by -x^{2}.

$$p=\frac{-x^{2}+2x-5}{-x^{2}}$$

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

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{5p-4}+1}{1-p}\text{; }x=\frac{-\sqrt{5p-4}+1}{1-p}\text{, }&p\neq 1\\x=\frac{5}{2}\text{, }&p=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{5p-4}+1}{1-p}\text{; }x=\frac{-\sqrt{5p-4}+1}{1-p}\text{, }&p\neq 1\text{ and }p\geq \frac{4}{5}\\x=\frac{5}{2}\text{, }&p=1\end{matrix}\right.$$