Solve 6=-2x-4x^2

$$6 = - 2 x - 4 x ^ { 2 }$$

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

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

$$-2x-4x^{2}=6$$

Subtract 6 from both sides.

$$-2x-4x^{2}-6=0$$

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.

$$-4x^{2}-2x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -2 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-4\right)\left(-6\right)}}{2\left(-4\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-4\right)\left(-6\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+16\left(-6\right)}}{2\left(-4\right)}$$

Multiply 16 times -6.

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

Add 4 to -96.

$$x=\frac{-\left(-2\right)±\sqrt{-92}}{2\left(-4\right)}$$

Take the square root of -92.

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

The opposite of -2 is 2.

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

Multiply 2 times -4.

$$x=\frac{2±2\sqrt{23}i}{-8}$$

Now solve the equation x=\frac{2±2\sqrt{23}i}{-8} when ± is plus. Add 2 to 2i\sqrt{23}.

$$x=\frac{2+2\sqrt{23}i}{-8}$$

Divide 2+2i\sqrt{23} by -8.

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

Now solve the equation x=\frac{2±2\sqrt{23}i}{-8} when ± is minus. Subtract 2i\sqrt{23} from 2.

$$x=\frac{-2\sqrt{23}i+2}{-8}$$

Divide 2-2i\sqrt{23} by -8.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$-2x-4x^{2}=6$$

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.

$$-4x^{2}-2x=6$$

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{2}{-4}\right)x=\frac{6}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

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

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 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.

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

Subtract \frac{1}{4} from both sides of the equation.

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