Solve 2=-1/2x^2+x+4

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

Solve for x

$$x=\sqrt{5}+1\approx 3.236067977$$

$$x=1-\sqrt{5}\approx -1.236067977$$

Steps Using the Quadratic Formula

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

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

Subtract 2 from both sides.

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

Subtract 2 from 4 to get 2.

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

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

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

Square 1.

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

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

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

Multiply 2 times 2.

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

Add 1 to 4.

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

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

$$x=\frac{-1±\sqrt{5}}{-1}$$

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

$$x=\frac{\sqrt{5}-1}{-1}$$

Divide -1+\sqrt{5} by -1.

$$x=1-\sqrt{5}$$

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

$$x=\frac{-\sqrt{5}-1}{-1}$$

Divide -1-\sqrt{5} by -1.

$$x=\sqrt{5}+1$$

The equation is now solved.

$$x=1-\sqrt{5} x=\sqrt{5}+1$$

Steps for Completing the Square

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

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

Subtract 4 from both sides.

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

Subtract 4 from 2 to get -2.

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

Multiply both sides by -2.

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

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

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

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

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

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

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

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.

$$x^{2}-2x+1=4+1$$

Add 4 to 1.

$$x^{2}-2x+1=5$$

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}.

$$\left(x-1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x-1=\sqrt{5} x-1=-\sqrt{5}$$

Add 1 to both sides of the equation.

$$x=\sqrt{5}+1 x=1-\sqrt{5}$$