Solve 12/(x+8)=5x

$$\frac { 12 } { ( x + 8 ) } = 5 x$$

Solve for x

$$x=\frac{2\sqrt{115}}{5}-4\approx 0.289522118$$

$$x=-\frac{2\sqrt{115}}{5}-4\approx -8.289522118$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -8 since division by zero is not defined. Multiply both sides of the equation by x+8.

$$12=5x\left(x+8\right)$$

Use the distributive property to multiply 5x by x+8.

$$12=5x^{2}+40x$$

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

$$5x^{2}+40x=12$$

Subtract 12 from both sides.

$$5x^{2}+40x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 40 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 5\left(-12\right)}}{2\times 5}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 5\left(-12\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-40±\sqrt{1600-20\left(-12\right)}}{2\times 5}$$

Multiply -20 times -12.

$$x=\frac{-40±\sqrt{1600+240}}{2\times 5}$$

Add 1600 to 240.

$$x=\frac{-40±\sqrt{1840}}{2\times 5}$$

Take the square root of 1840.

$$x=\frac{-40±4\sqrt{115}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-40±4\sqrt{115}}{10}$$

Now solve the equation x=\frac{-40±4\sqrt{115}}{10} when ± is plus. Add -40 to 4\sqrt{115}.

$$x=\frac{4\sqrt{115}-40}{10}$$

Divide -40+4\sqrt{115} by 10.

$$x=\frac{2\sqrt{115}}{5}-4$$

Now solve the equation x=\frac{-40±4\sqrt{115}}{10} when ± is minus. Subtract 4\sqrt{115} from -40.

$$x=\frac{-4\sqrt{115}-40}{10}$$

Divide -40-4\sqrt{115} by 10.

$$x=-\frac{2\sqrt{115}}{5}-4$$

The equation is now solved.

$$x=\frac{2\sqrt{115}}{5}-4 x=-\frac{2\sqrt{115}}{5}-4$$

Steps for Completing the Square

Variable x cannot be equal to -8 since division by zero is not defined. Multiply both sides of the equation by x+8.

$$12=5x\left(x+8\right)$$

Use the distributive property to multiply 5x by x+8.

$$12=5x^{2}+40x$$

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

$$5x^{2}+40x=12$$

Divide both sides by 5.

$$\frac{5x^{2}+40x}{5}=\frac{12}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{40}{5}x=\frac{12}{5}$$

Divide 40 by 5.

$$x^{2}+8x=\frac{12}{5}$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=\frac{12}{5}+4^{2}$$

Square 4.

$$x^{2}+8x+16=\frac{12}{5}+16$$

Add \frac{12}{5} to 16.

$$x^{2}+8x+16=\frac{92}{5}$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=\frac{92}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{\frac{92}{5}}$$

Simplify.

$$x+4=\frac{2\sqrt{115}}{5} x+4=-\frac{2\sqrt{115}}{5}$$

Subtract 4 from both sides of the equation.

$$x=\frac{2\sqrt{115}}{5}-4 x=-\frac{2\sqrt{115}}{5}-4$$