Solve (x-2)+1=2(x^2)

$$( x - 2 ) + 1 = 2 ( x ^ { 2 } )$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{7}i+1}{4}\approx 0.25-0.661437828i$$

$$x=\frac{1+\sqrt{7}i}{4}\approx 0.25+0.661437828i$$

Steps Using the Quadratic Formula

Add -2 and 1 to get -1.

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

Subtract 2x^{2} from both sides.

$$x-1-2x^{2}=0$$

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.

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

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

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

Square 1.

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

Multiply -4 times -2.

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

Multiply 8 times -1.

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

Add 1 to -8.

$$x=\frac{-1±\sqrt{-7}}{2\left(-2\right)}$$

Take the square root of -7.

$$x=\frac{-1±\sqrt{7}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-1±\sqrt{7}i}{-4}$$

Now solve the equation x=\frac{-1±\sqrt{7}i}{-4} when ± is plus. Add -1 to i\sqrt{7}.

$$x=\frac{-1+\sqrt{7}i}{-4}$$

Divide -1+i\sqrt{7} by -4.

$$x=\frac{-\sqrt{7}i+1}{4}$$

Now solve the equation x=\frac{-1±\sqrt{7}i}{-4} when ± is minus. Subtract i\sqrt{7} from -1.

$$x=\frac{-\sqrt{7}i-1}{-4}$$

Divide -1-i\sqrt{7} by -4.

$$x=\frac{1+\sqrt{7}i}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{7}i+1}{4} x=\frac{1+\sqrt{7}i}{4}$$

Steps for Completing the Square

Add -2 and 1 to get -1.

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

Subtract 2x^{2} from both sides.

$$x-1-2x^{2}=0$$

Add 1 to both sides. Anything plus zero gives itself.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 1 by -2.

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

Divide 1 by -2.

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

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.

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

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

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

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{7}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{7}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{7}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{7}i}{4} x-\frac{1}{4}=-\frac{\sqrt{7}i}{4}$$

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

$$x=\frac{1+\sqrt{7}i}{4} x=\frac{-\sqrt{7}i+1}{4}$$