Solve left(2*xright)^2+x+1=0

$${ \left(2 \times x \right) }^{ 2 } +x+1 = 0$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{15}i}{8}\approx -0.125+0.484122918i$$

$$x=\frac{-\sqrt{15}i-1}{8}\approx -0.125-0.484122918i$$

Steps Using the Quadratic Formula

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}+x+1=0$$

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

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

Square 1.

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

Multiply -4 times 4.

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

Add 1 to -16.

$$x=\frac{-1±\sqrt{-15}}{2\times 4}$$

Take the square root of -15.

$$x=\frac{-1±\sqrt{15}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-1±\sqrt{15}i}{8}$$

Now solve the equation x=\frac{-1±\sqrt{15}i}{8} when ± is plus. Add -1 to i\sqrt{15}.

$$x=\frac{-1+\sqrt{15}i}{8}$$

Now solve the equation x=\frac{-1±\sqrt{15}i}{8} when ± is minus. Subtract i\sqrt{15} from -1.

$$x=\frac{-\sqrt{15}i-1}{8}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{15}i}{8} x=\frac{-\sqrt{15}i-1}{8}$$

Steps for Completing the Square

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$4x^{2}+x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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.

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

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

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

Add -\frac{1}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{15}{64}$$

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}.

$$\left(x+\frac{1}{8}\right)^{2}=-\frac{15}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{-\frac{15}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{\sqrt{15}i}{8} x+\frac{1}{8}=-\frac{\sqrt{15}i}{8}$$

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

$$x=\frac{-1+\sqrt{15}i}{8} x=\frac{-\sqrt{15}i-1}{8}$$