Solve x^2+2x-4-3=0

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

Solve for x

$$x=2\sqrt{2}-1\approx 1.828427125$$

$$x=-2\sqrt{2}-1\approx -3.828427125$$

Steps Using the Quadratic Formula

Subtract 3 from -4 to get -7.

$$x^{2}+2x-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2.

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

Multiply -4 times -7.

$$x=\frac{-2±\sqrt{4+28}}{2}$$

Add 4 to 28.

$$x=\frac{-2±\sqrt{32}}{2}$$

Take the square root of 32.

$$x=\frac{-2±4\sqrt{2}}{2}$$

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

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

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

$$x=2\sqrt{2}-1$$

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

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

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

$$x=-2\sqrt{2}-1$$

The equation is now solved.

$$x=2\sqrt{2}-1 x=-2\sqrt{2}-1$$

Steps for Completing the Square

Subtract 3 from -4 to get -7.

$$x^{2}+2x-7=0$$

Add 7 to both sides. Anything plus zero gives itself.

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

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

$$x^{2}+2x+1^{2}=7+1^{2}$$

Square 1.

$$x^{2}+2x+1=7+1$$

Add 7 to 1.

$$x^{2}+2x+1=8$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=2\sqrt{2} x+1=-2\sqrt{2}$$

Subtract 1 from both sides of the equation.

$$x=2\sqrt{2}-1 x=-2\sqrt{2}-1$$