Solve -2x^2-2x+12/4-x^2=0

$$- \frac { 2 x ^ { 2 } - 2 x + 12 } { 4 - x ^ { 2 } } = 0$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{23}i}{2}\approx 0.5+2.397915762i$$

$$x=\frac{-\sqrt{23}i+1}{2}\approx 0.5-2.397915762i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(-x-2\right).

$$-\left(2x^{2}-2x+12\right)=0$$

To find the opposite of 2x^{2}-2x+12, find the opposite of each term.

$$-2x^{2}+2x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 2 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-2±\sqrt{4+8\left(-12\right)}}{2\left(-2\right)}$$

Multiply 8 times -12.

$$x=\frac{-2±\sqrt{4-96}}{2\left(-2\right)}$$

Add 4 to -96.

$$x=\frac{-2±\sqrt{-92}}{2\left(-2\right)}$$

Take the square root of -92.

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

Multiply 2 times -2.

$$x=\frac{-2±2\sqrt{23}i}{-4}$$

Now solve the equation x=\frac{-2±2\sqrt{23}i}{-4} when ± is plus. Add -2 to 2i\sqrt{23}.

$$x=\frac{-2+2\sqrt{23}i}{-4}$$

Divide -2+2i\sqrt{23} by -4.

$$x=\frac{-\sqrt{23}i+1}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{23}i}{-4} when ± is minus. Subtract 2i\sqrt{23} from -2.

$$x=\frac{-2\sqrt{23}i-2}{-4}$$

Divide -2-2i\sqrt{23} by -4.

$$x=\frac{1+\sqrt{23}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{23}i+1}{2} x=\frac{1+\sqrt{23}i}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(-x-2\right).

$$-\left(2x^{2}-2x+12\right)=0$$

To find the opposite of 2x^{2}-2x+12, find the opposite of each term.

$$-2x^{2}+2x-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

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

Divide both sides by -2.

$$\frac{-2x^{2}+2x}{-2}=\frac{12}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{2}{-2}x=\frac{12}{-2}$$

Divide 2 by -2.

$$x^{2}-x=\frac{12}{-2}$$

Divide 12 by -2.

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

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

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-6+\frac{1}{4}$$

Add -6 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=-\frac{23}{4}$$

Factor x^{2}-x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{23}i}{2} x-\frac{1}{2}=-\frac{\sqrt{23}i}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{1+\sqrt{23}i}{2} x=\frac{-\sqrt{23}i+1}{2}$$