Solve -2=4x^2+x

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{31}i-1}{8}\approx -0.125-0.695970545i$$

$$x=\frac{-1+\sqrt{31}i}{8}\approx -0.125+0.695970545i$$

Steps Using the Quadratic Formula

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

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

Add 2 to both sides.

$$4x^{2}+x+2=0$$

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

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\times 2}}{2\times 4}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 4\times 2}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-1±\sqrt{1-16\times 2}}{2\times 4}$$

Multiply -16 times 2.

$$x=\frac{-1±\sqrt{1-32}}{2\times 4}$$

Add 1 to -32.

$$x=\frac{-1±\sqrt{-31}}{2\times 4}$$

Take the square root of -31.

$$x=\frac{-1±\sqrt{31}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-1±\sqrt{31}i}{8}$$

Now solve the equation x=\frac{-1±\sqrt{31}i}{8} when ± is plus. Add -1 to i\sqrt{31}.

$$x=\frac{-1+\sqrt{31}i}{8}$$

Now solve the equation x=\frac{-1±\sqrt{31}i}{8} when ± is minus. Subtract i\sqrt{31} from -1.

$$x=\frac{-\sqrt{31}i-1}{8}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{31}i}{8} x=\frac{-\sqrt{31}i-1}{8}$$

Steps for Completing the Square

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

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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

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

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=-\frac{1}{2}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{1}{2}+\frac{1}{64}$$

Add -\frac{1}{2} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{31}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=-\frac{31}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{-\frac{31}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{\sqrt{31}i}{8} x+\frac{1}{8}=-\frac{\sqrt{31}i}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=\frac{-1+\sqrt{31}i}{8} x=\frac{-\sqrt{31}i-1}{8}$$