Solve 6=-0.5x^2+2x+6

$$6=-0.5 { x }^{ 2 } +2x+6$$

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

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

$$-0.5x^{2}+2x+6=6$$

Subtract 6 from both sides.

$$-0.5x^{2}+2x+6-6=0$$

Subtract 6 from 6 to get 0.

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

Factor out x.

$$x\left(-0.5x+2\right)=0$$

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

$$x=0 x=4$$

Steps Using the Quadratic Formula

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

$$-0.5x^{2}+2x+6=6$$

Subtract 6 from both sides.

$$-0.5x^{2}+2x+6-6=0$$

Subtract 6 from 6 to get 0.

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

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

$$x=\frac{-2±\sqrt{2^{2}}}{2\left(-0.5\right)}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\left(-0.5\right)}$$

Multiply 2 times -0.5.

$$x=\frac{-2±2}{-1}$$

Now solve the equation x=\frac{-2±2}{-1} when ± is plus. Add -2 to 2.

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

Divide 0 by -1.

$$x=0$$

Now solve the equation x=\frac{-2±2}{-1} when ± is minus. Subtract 2 from -2.

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

Divide -4 by -1.

$$x=4$$

The equation is now solved.

$$x=0 x=4$$

Steps for Completing the Square

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

$$-0.5x^{2}+2x+6=6$$

Subtract 6 from both sides.

$$-0.5x^{2}+2x=6-6$$

Subtract 6 from 6 to get 0.

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

Multiply both sides by -2.

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

Dividing by -0.5 undoes the multiplication by -0.5.

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

Divide 2 by -0.5 by multiplying 2 by the reciprocal of -0.5.

$$x^{2}-4x=\frac{0}{-0.5}$$

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

$$x^{2}-4x=0$$

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

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

Square -2.

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

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-2=2 x-2=-2$$

Add 2 to both sides of the equation.

$$x=4 x=0$$