Solve E(x)=5quadE[x(x-1)]=27

$$E ( x ) = 5 \quad E [ x ( x - 1 ) ] = 27$$

Solve for E

$$E=\frac{27}{5x\left(x-1\right)}$$

$$x\neq 1\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 5Ex by x-1.

$$5Ex^{2}-5Ex=27$$

Combine all terms containing E.

$$\left(5x^{2}-5x\right)E=27$$

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

$$\frac{\left(5x^{2}-5x\right)E}{5x^{2}-5x}=\frac{27}{5x^{2}-5x}$$

Dividing by 5x^{2}-5x undoes the multiplication by 5x^{2}-5x.

$$E=\frac{27}{5x^{2}-5x}$$

Divide 27 by 5x^{2}-5x.

$$E=\frac{27}{5x\left(x-1\right)}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{25E^{2}+540E}}{10E}+\frac{1}{2}$$

$$x=-\frac{\sqrt{25E^{2}+540E}}{10E}+\frac{1}{2}\text{, }E\neq 0$$

Solve for x

$$x=\frac{\sqrt{25E^{2}+540E}}{10E}+\frac{1}{2}$$

$$x=-\frac{\sqrt{25E^{2}+540E}}{10E}+\frac{1}{2}\text{, }E>0\text{ or }E\leq -\frac{108}{5}$$