Solve 32=5x*(8x+16)

$$32=5x \times (8x+16)$$

Solve for x

$$x=\frac{3\sqrt{5}}{5}-1\approx 0.341640786$$

$$x=-\frac{3\sqrt{5}}{5}-1\approx -2.341640786$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x by 8x+16.

$$32=40x^{2}+80x$$

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

$$40x^{2}+80x=32$$

Subtract 32 from both sides.

$$40x^{2}+80x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, 80 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-80±\sqrt{80^{2}-4\times 40\left(-32\right)}}{2\times 40}$$

Square 80.

$$x=\frac{-80±\sqrt{6400-4\times 40\left(-32\right)}}{2\times 40}$$

Multiply -4 times 40.

$$x=\frac{-80±\sqrt{6400-160\left(-32\right)}}{2\times 40}$$

Multiply -160 times -32.

$$x=\frac{-80±\sqrt{6400+5120}}{2\times 40}$$

Add 6400 to 5120.

$$x=\frac{-80±\sqrt{11520}}{2\times 40}$$

Take the square root of 11520.

$$x=\frac{-80±48\sqrt{5}}{2\times 40}$$

Multiply 2 times 40.

$$x=\frac{-80±48\sqrt{5}}{80}$$

Now solve the equation x=\frac{-80±48\sqrt{5}}{80} when ± is plus. Add -80 to 48\sqrt{5}.

$$x=\frac{48\sqrt{5}-80}{80}$$

Divide -80+48\sqrt{5} by 80.

$$x=\frac{3\sqrt{5}}{5}-1$$

Now solve the equation x=\frac{-80±48\sqrt{5}}{80} when ± is minus. Subtract 48\sqrt{5} from -80.

$$x=\frac{-48\sqrt{5}-80}{80}$$

Divide -80-48\sqrt{5} by 80.

$$x=-\frac{3\sqrt{5}}{5}-1$$

The equation is now solved.

$$x=\frac{3\sqrt{5}}{5}-1 x=-\frac{3\sqrt{5}}{5}-1$$

Steps for Completing the Square

Use the distributive property to multiply 5x by 8x+16.

$$32=40x^{2}+80x$$

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

$$40x^{2}+80x=32$$

Divide both sides by 40.

$$\frac{40x^{2}+80x}{40}=\frac{32}{40}$$

Dividing by 40 undoes the multiplication by 40.

$$x^{2}+\frac{80}{40}x=\frac{32}{40}$$

Divide 80 by 40.

$$x^{2}+2x=\frac{32}{40}$$

Reduce the fraction \frac{32}{40} to lowest terms by extracting and canceling out 8.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{4}{5}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{4}{5}+1$$

Add \frac{4}{5} to 1.

$$x^{2}+2x+1=\frac{9}{5}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{9}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{9}{5}}$$

Simplify.

$$x+1=\frac{3\sqrt{5}}{5} x+1=-\frac{3\sqrt{5}}{5}$$

Subtract 1 from both sides of the equation.

$$x=\frac{3\sqrt{5}}{5}-1 x=-\frac{3\sqrt{5}}{5}-1$$