Solve 4x^2=-8x

$$4 { x }^{ 2 } =-8x$$

Solve for x

$$x=-2$$

$$x=0$$

Steps Using Factoring

Add 8x to both sides.

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

Factor out x.

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

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

$$x=0 x=-2$$

Steps Using the Quadratic Formula

Add 8x to both sides.

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

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

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

Take the square root of 8^{2}.

$$x=\frac{-8±8}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-8±8}{8}$$

Now solve the equation x=\frac{-8±8}{8} when ± is plus. Add -8 to 8.

$$x=\frac{0}{8}$$

Divide 0 by 8.

$$x=0$$

Now solve the equation x=\frac{-8±8}{8} when ± is minus. Subtract 8 from -8.

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

Divide -16 by 8.

$$x=-2$$

The equation is now solved.

$$x=0 x=-2$$

Steps for Completing the Square

Add 8x to both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide 8 by 4.

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

Divide 0 by 4.

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

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.

$$x^{2}+2x+1^{2}=1^{2}$$

Square 1.

$$x^{2}+2x+1=1$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$