Solve -x+3=-1/2x^2+x-3/2

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

Solve for x (complex solution)

$$x=2+\sqrt{5}i\approx 2+2.236067977i$$

$$x=-\sqrt{5}i+2\approx 2-2.236067977i$$

Steps Using the Quadratic Formula

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

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

Subtract x from both sides.

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

Add \frac{3}{2} to both sides.

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

Add 3 and \frac{3}{2} to get \frac{9}{2}.

$$-x+\frac{9}{2}+\frac{1}{2}x^{2}-x=0$$

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

$$-2x+\frac{9}{2}+\frac{1}{2}x^{2}=0$$

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.

$$\frac{1}{2}x^{2}-2x+\frac{9}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -2 for b, and \frac{9}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

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

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

Multiply -2 times \frac{9}{2}.

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

Add 4 to -9.

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

Take the square root of -5.

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

The opposite of -2 is 2.

$$x=\frac{2±\sqrt{5}i}{2\times \frac{1}{2}}$$

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

$$x=\frac{2±\sqrt{5}i}{1}$$

Now solve the equation x=\frac{2±\sqrt{5}i}{1} when ± is plus. Add 2 to i\sqrt{5}.

$$x=\frac{2+\sqrt{5}i}{1}$$

Divide 2+i\sqrt{5} by 1.

$$x=2+\sqrt{5}i$$

Now solve the equation x=\frac{2±\sqrt{5}i}{1} when ± is minus. Subtract i\sqrt{5} from 2.

$$x=\frac{-\sqrt{5}i+2}{1}$$

Divide 2-i\sqrt{5} by 1.

$$x=-\sqrt{5}i+2$$

The equation is now solved.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$

Steps for Completing the Square

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

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

Subtract x from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from -\frac{3}{2} to get -\frac{9}{2}.

$$-x+\frac{1}{2}x^{2}-x=-\frac{9}{2}$$

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

$$-2x+\frac{1}{2}x^{2}=-\frac{9}{2}$$

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.

$$\frac{1}{2}x^{2}-2x=-\frac{9}{2}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-2x}{\frac{1}{2}}=-\frac{\frac{9}{2}}{\frac{1}{2}}$$

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

$$x^{2}+\left(-\frac{2}{\frac{1}{2}}\right)x=-\frac{\frac{9}{2}}{\frac{1}{2}}$$

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

$$x^{2}-4x=-\frac{\frac{9}{2}}{\frac{1}{2}}$$

Divide -\frac{9}{2} by \frac{1}{2} by multiplying -\frac{9}{2} by the reciprocal of \frac{1}{2}.

$$x^{2}-4x=-9$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-9+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=-9+4$$

Add -9 to 4.

$$x^{2}-4x+4=-5$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=\sqrt{5}i x-2=-\sqrt{5}i$$

Add 2 to both sides of the equation.

$$x=2+\sqrt{5}i x=-\sqrt{5}i+2$$