Solve -x=-1/2x^2+2x+6

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

Solve for x

$$x=\sqrt{21}+3\approx 7.582575695$$

$$x=3-\sqrt{21}\approx -1.582575695$$

Steps Using the Quadratic Formula

Add \frac{1}{2}x^{2} to both sides.

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

Subtract 2x from both sides.

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

Subtract 6 from both sides.

$$-x+\frac{1}{2}x^{2}-2x-6=0$$

Combine -x and -2x to get -3x.

$$-3x+\frac{1}{2}x^{2}-6=0$$

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.

$$\frac{1}{2}x^{2}-3x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -3 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times \frac{1}{2}\left(-6\right)}}{2\times \frac{1}{2}}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times \frac{1}{2}\left(-6\right)}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-\left(-3\right)±\sqrt{9-2\left(-6\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+12}}{2\times \frac{1}{2}}$$

Add 9 to 12.

$$x=\frac{-\left(-3\right)±\sqrt{21}}{2\times \frac{1}{2}}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{21}}{2\times \frac{1}{2}}$$

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

$$x=\frac{3±\sqrt{21}}{1}$$

Now solve the equation x=\frac{3±\sqrt{21}}{1} when ± is plus. Add 3 to \sqrt{21}.

$$x=\frac{\sqrt{21}+3}{1}$$

Divide 3+\sqrt{21} by 1.

$$x=\sqrt{21}+3$$

Now solve the equation x=\frac{3±\sqrt{21}}{1} when ± is minus. Subtract \sqrt{21} from 3.

$$x=\frac{3-\sqrt{21}}{1}$$

Divide 3-\sqrt{21} by 1.

$$x=3-\sqrt{21}$$

The equation is now solved.

$$x=\sqrt{21}+3 x=3-\sqrt{21}$$

Steps for Completing the Square

Add \frac{1}{2}x^{2} to both sides.

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

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

$$-3x+\frac{1}{2}x^{2}=6$$

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.

$$\frac{1}{2}x^{2}-3x=6$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-3x}{\frac{1}{2}}=\frac{6}{\frac{1}{2}}$$

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

$$x^{2}+\left(-\frac{3}{\frac{1}{2}}\right)x=\frac{6}{\frac{1}{2}}$$

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

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

Divide 6 by \frac{1}{2} by multiplying 6 by the reciprocal of \frac{1}{2}.

$$x^{2}-6x=12$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=12+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=12+9$$

Add 12 to 9.

$$x^{2}-6x+9=21$$

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}.

$$\left(x-3\right)^{2}=21$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{21}$$

Simplify.

$$x-3=\sqrt{21} x-3=-\sqrt{21}$$

Add 3 to both sides of the equation.

$$x=\sqrt{21}+3 x=3-\sqrt{21}$$