Solve left(2xright)^2-2x-3=0

$${ \left(2x \right) }^{ 2 } -2x-3=0$$

Solve for x

$$x = \frac{\sqrt{13} + 1}{4} \approx 1.151387819$$

$$x=\frac{1-\sqrt{13}}{4}\approx -0.651387819$$

Steps Using the Quadratic Formula

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

$$2^{2}x^{2}-2x-3=0$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-2x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 4\left(-3\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-2\right)±\sqrt{4-16\left(-3\right)}}{2\times 4}$$

Multiply -16 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+48}}{2\times 4}$$

Add 4 to 48.

$$x=\frac{-\left(-2\right)±\sqrt{52}}{2\times 4}$$

Take the square root of 52.

$$x=\frac{-\left(-2\right)±2\sqrt{13}}{2\times 4}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{13}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{2±2\sqrt{13}}{8}$$

Now solve the equation x=\frac{2±2\sqrt{13}}{8} when ± is plus. Add 2 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}+2}{8}$$

Divide 2+2\sqrt{13} by 8.

$$x=\frac{\sqrt{13}+1}{4}$$

Now solve the equation x=\frac{2±2\sqrt{13}}{8} when ± is minus. Subtract 2\sqrt{13} from 2.

$$x=\frac{2-2\sqrt{13}}{8}$$

Divide 2-2\sqrt{13} by 8.

$$x=\frac{1-\sqrt{13}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{13}+1}{4} x=\frac{1-\sqrt{13}}{4}$$

Steps for Completing the Square

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

$$2^{2}x^{2}-2x-3=0$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-2x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$4x^{2}-2x=3$$

Divide both sides by 4.

$$\frac{4x^{2}-2x}{4}=\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{2}{4}\right)x=\frac{3}{4}$$

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

$$x^{2}-\frac{1}{2}x=\frac{3}{4}$$

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.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{3}{4}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{3}{4}+\frac{1}{16}$$

Add \frac{3}{4} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{13}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{13}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{13}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{13}}{4} x-\frac{1}{4}=-\frac{\sqrt{13}}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{\sqrt{13}+1}{4} x=\frac{1-\sqrt{13}}{4}$$