Solve x=-0.5x^2

$$x=-0.5 { x }^{ 2 }$$

Solve for x

$$x=-2$$

$$x=0$$

Steps Using Factoring

Add 0.5x^{2} to both sides.

$$x+0.5x^{2}=0$$

Factor out x.

$$x\left(1+0.5x\right)=0$$

To find equation solutions, solve x=0 and 1+\frac{x}{2}=0.

$$x=0 x=-2$$

Steps Using the Quadratic Formula

Add 0.5x^{2} to both sides.

$$x+0.5x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$0.5x^{2}+x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.5 for a, 1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}}}{2\times 0.5}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\times 0.5}$$

Multiply 2 times 0.5.

$$x=\frac{-1±1}{1}$$

Now solve the equation x=\frac{-1±1}{1} when ± is plus. Add -1 to 1.

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

Divide 0 by 1.

$$x=0$$

Now solve the equation x=\frac{-1±1}{1} when ± is minus. Subtract 1 from -1.

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

Divide -2 by 1.

$$x=-2$$

The equation is now solved.

$$x=0 x=-2$$

Steps for Completing the Square

Add 0.5x^{2} to both sides.

$$x+0.5x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$0.5x^{2}+x=0$$

Multiply both sides by 2.

$$\frac{0.5x^{2}+x}{0.5}=\frac{0}{0.5}$$

Dividing by 0.5 undoes the multiplication by 0.5.

$$x^{2}+\frac{1}{0.5}x=\frac{0}{0.5}$$

Divide 1 by 0.5 by multiplying 1 by the reciprocal of 0.5.

$$x^{2}+2x=\frac{0}{0.5}$$

Divide 0 by 0.5 by multiplying 0 by the reciprocal of 0.5.

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

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.

$$x=0 x=-2$$