Solve 4x(2x+1)=0

$$4 x ( 2 x + 1 ) = 0$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply 4x by 2x+1.

$$8x^{2}+4x=0$$

Factor out x.

$$x\left(8x+4\right)=0$$

To find equation solutions, solve x=0 and 8x+4=0.

$$x=0 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4x by 2x+1.

$$8x^{2}+4x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}}}{2\times 8}$$

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-4±4}{16}$$

Now solve the equation x=\frac{-4±4}{16} when ± is plus. Add -4 to 4.

$$x=\frac{0}{16}$$

Divide 0 by 16.

$$x=0$$

Now solve the equation x=\frac{-4±4}{16} when ± is minus. Subtract 4 from -4.

$$x=-\frac{8}{16}$$

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=0 x=-\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 4x by 2x+1.

$$8x^{2}+4x=0$$

Divide both sides by 8.

$$\frac{8x^{2}+4x}{8}=\frac{0}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{4}{8}x=\frac{0}{8}$$

Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{1}{2}x=\frac{0}{8}$$

Divide 0 by 8.

$$x^{2}+\frac{1}{2}x=0$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{1}{4} x+\frac{1}{4}=-\frac{1}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=0 x=-\frac{1}{2}$$