Solve 0=x^2-4x-3

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

Solve for x

$$x=\sqrt{7}+2\approx 4.645751311$$

$$x=2-\sqrt{7}\approx -0.645751311$$

Steps Using the Quadratic Formula

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

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

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

Multiply -4 times -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+12}}{2}$$

Add 16 to 12.

$$x=\frac{-\left(-4\right)±\sqrt{28}}{2}$$

Take the square root of 28.

$$x=\frac{-\left(-4\right)±2\sqrt{7}}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{7}}{2}$$

Now solve the equation x=\frac{4±2\sqrt{7}}{2} when ± is plus. Add 4 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+4}{2}$$

Divide 4+2\sqrt{7} by 2.

$$x=\sqrt{7}+2$$

Now solve the equation x=\frac{4±2\sqrt{7}}{2} when ± is minus. Subtract 2\sqrt{7} from 4.

$$x=\frac{4-2\sqrt{7}}{2}$$

Divide 4-2\sqrt{7} by 2.

$$x=2-\sqrt{7}$$

The equation is now solved.

$$x=\sqrt{7}+2 x=2-\sqrt{7}$$

Steps for Completing the Square

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

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

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

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add 3 to 4.

$$x^{2}-4x+4=7$$

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

$$\left(x-2\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{7}$$

Simplify.

$$x-2=\sqrt{7} x-2=-\sqrt{7}$$

Add 2 to both sides of the equation.

$$x=\sqrt{7}+2 x=2-\sqrt{7}$$