Solve -x+3=-2x^2

$$- x + 3 = - 2 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{23}i}{4}\approx 0.25+1.198957881i$$

$$x=\frac{-\sqrt{23}i+1}{4}\approx 0.25-1.198957881i$$

Steps Using the Quadratic Formula

Add 2x^{2} to both sides.

$$-x+3+2x^{2}=0$$

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.

$$2x^{2}-x+3=0$$

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

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

Multiply -4 times 2.

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

Multiply -8 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-24}}{2\times 2}$$

Add 1 to -24.

$$x=\frac{-\left(-1\right)±\sqrt{-23}}{2\times 2}$$

Take the square root of -23.

$$x=\frac{-\left(-1\right)±\sqrt{23}i}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{23}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{23}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{23}i}{4} when ± is plus. Add 1 to i\sqrt{23}.

$$x=\frac{1+\sqrt{23}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{23}i}{4} when ± is minus. Subtract i\sqrt{23} from 1.

$$x=\frac{-\sqrt{23}i+1}{4}$$

The equation is now solved.

$$x=\frac{1+\sqrt{23}i}{4} x=\frac{-\sqrt{23}i+1}{4}$$

Steps for Completing the Square

Add 2x^{2} to both sides.

$$-x+3+2x^{2}=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$-x+2x^{2}=-3$$

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.

$$2x^{2}-x=-3$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{3}{2}+\frac{1}{16}$$

Add -\frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{23}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{23}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{23}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{23}i}{4} x-\frac{1}{4}=-\frac{\sqrt{23}i}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1+\sqrt{23}i}{4} x=\frac{-\sqrt{23}i+1}{4}$$