Solve -1/2left(2-xright)^2+4+2x+6=x

$$- \frac{ 1 }{ 2 } { \left(2-x \right) }^{ 2 } +4+2x+6=x$$

Solve for x

$$x=-2$$

$$x=8$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.

$$-\frac{1}{2}\left(4-4x+x^{2}\right)+4+2x+6=x$$

Use the distributive property to multiply -\frac{1}{2} by 4-4x+x^{2}.

$$-2+2x-\frac{1}{2}x^{2}+4+2x+6=x$$

Add -2 and 4 to get 2.

$$2+2x-\frac{1}{2}x^{2}+2x+6=x$$

Combine 2x and 2x to get 4x.

$$2+4x-\frac{1}{2}x^{2}+6=x$$

Add 2 and 6 to get 8.

$$8+4x-\frac{1}{2}x^{2}=x$$

Subtract x from both sides.

$$8+4x-\frac{1}{2}x^{2}-x=0$$

Combine 4x and -x to get 3x.

$$8+3x-\frac{1}{2}x^{2}=0$$

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.

$$-\frac{1}{2}x^{2}+3x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, 3 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-\frac{1}{2}\right)\times 8}}{2\left(-\frac{1}{2}\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-\frac{1}{2}\right)\times 8}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{-3±\sqrt{9+2\times 8}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times 8.

$$x=\frac{-3±\sqrt{9+16}}{2\left(-\frac{1}{2}\right)}$$

Add 9 to 16.

$$x=\frac{-3±\sqrt{25}}{2\left(-\frac{1}{2}\right)}$$

Take the square root of 25.

$$x=\frac{-3±5}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{-3±5}{-1}$$

Now solve the equation x=\frac{-3±5}{-1} when ± is plus. Add -3 to 5.

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

Divide 2 by -1.

$$x=-2$$

Now solve the equation x=\frac{-3±5}{-1} when ± is minus. Subtract 5 from -3.

$$x=-\frac{8}{-1}$$

Divide -8 by -1.

$$x=8$$

The equation is now solved.

$$x=-2 x=8$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.

$$-\frac{1}{2}\left(4-4x+x^{2}\right)+4+2x+6=x$$

Use the distributive property to multiply -\frac{1}{2} by 4-4x+x^{2}.

$$-2+2x-\frac{1}{2}x^{2}+4+2x+6=x$$

Add -2 and 4 to get 2.

$$2+2x-\frac{1}{2}x^{2}+2x+6=x$$

Combine 2x and 2x to get 4x.

$$2+4x-\frac{1}{2}x^{2}+6=x$$

Add 2 and 6 to get 8.

$$8+4x-\frac{1}{2}x^{2}=x$$

Subtract x from both sides.

$$8+4x-\frac{1}{2}x^{2}-x=0$$

Combine 4x and -x to get 3x.

$$8+3x-\frac{1}{2}x^{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$3x-\frac{1}{2}x^{2}=-8$$

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.

$$-\frac{1}{2}x^{2}+3x=-8$$

Multiply both sides by -2.

$$\frac{-\frac{1}{2}x^{2}+3x}{-\frac{1}{2}}=-\frac{8}{-\frac{1}{2}}$$

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

$$x^{2}+\frac{3}{-\frac{1}{2}}x=-\frac{8}{-\frac{1}{2}}$$

Divide 3 by -\frac{1}{2} by multiplying 3 by the reciprocal of -\frac{1}{2}.

$$x^{2}-6x=-\frac{8}{-\frac{1}{2}}$$

Divide -8 by -\frac{1}{2} by multiplying -8 by the reciprocal of -\frac{1}{2}.

$$x^{2}-6x=16$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=16+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=16+9$$

Add 16 to 9.

$$x^{2}-6x+9=25$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-3=5 x-3=-5$$

Add 3 to both sides of the equation.

$$x=8 x=-2$$