Solve left(4xright)^2-2x+6=0

$${ \left(4x \right) }^{ 2 } -2x+6 = 0$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{95}i}{16}\approx 0.0625+0.609174647i$$

$$x=\frac{-\sqrt{95}i+1}{16}\approx 0.0625-0.609174647i$$

Steps Using the Quadratic Formula

Expand \left(4x\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-2x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -2 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times 16.

$$x=\frac{-\left(-2\right)±\sqrt{4-64\times 6}}{2\times 16}$$

Multiply -64 times 6.

$$x=\frac{-\left(-2\right)±\sqrt{4-384}}{2\times 16}$$

Add 4 to -384.

$$x=\frac{-\left(-2\right)±\sqrt{-380}}{2\times 16}$$

Take the square root of -380.

$$x=\frac{-\left(-2\right)±2\sqrt{95}i}{2\times 16}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{95}i}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{2±2\sqrt{95}i}{32}$$

Now solve the equation x=\frac{2±2\sqrt{95}i}{32} when ± is plus. Add 2 to 2i\sqrt{95}.

$$x=\frac{2+2\sqrt{95}i}{32}$$

Divide 2+2i\sqrt{95} by 32.

$$x=\frac{1+\sqrt{95}i}{16}$$

Now solve the equation x=\frac{2±2\sqrt{95}i}{32} when ± is minus. Subtract 2i\sqrt{95} from 2.

$$x=\frac{-2\sqrt{95}i+2}{32}$$

Divide 2-2i\sqrt{95} by 32.

$$x=\frac{-\sqrt{95}i+1}{16}$$

The equation is now solved.

$$x=\frac{1+\sqrt{95}i}{16} x=\frac{-\sqrt{95}i+1}{16}$$

Steps for Completing the Square

Expand \left(4x\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-2x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$16x^{2}-2x=-6$$

Divide both sides by 16.

$$\frac{16x^{2}-2x}{16}=-\frac{6}{16}$$

Dividing by 16 undoes the multiplication by 16.

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

Reduce the fraction \frac{-2}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{8}x=-\frac{6}{16}$$

Reduce the fraction \frac{-6}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{8}x=-\frac{3}{8}$$

Divide -\frac{1}{8}, the coefficient of the x term, by 2 to get -\frac{1}{16}. Then add the square of -\frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{8}x+\left(-\frac{1}{16}\right)^{2}=-\frac{3}{8}+\left(-\frac{1}{16}\right)^{2}$$

Square -\frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=-\frac{3}{8}+\frac{1}{256}$$

Add -\frac{3}{8} to \frac{1}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=-\frac{95}{256}$$

Factor x^{2}-\frac{1}{8}x+\frac{1}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{16}\right)^{2}=-\frac{95}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{16}\right)^{2}}=\sqrt{-\frac{95}{256}}$$

Simplify.

$$x-\frac{1}{16}=\frac{\sqrt{95}i}{16} x-\frac{1}{16}=-\frac{\sqrt{95}i}{16}$$

Add \frac{1}{16} to both sides of the equation.

$$x=\frac{1+\sqrt{95}i}{16} x=\frac{-\sqrt{95}i+1}{16}$$