Solve (x+3)(x+1)=2x+7

$$( x + 3 ) ( x + 1 ) = 2 x + 7$$

Solve for x (complex solution)

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\left(\sqrt{5}+1\right)\approx -3.236067977$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+3 by x+1 and combine like terms.

$$x^{2}+4x+3=2x+7$$

Subtract 2x from both sides.

$$x^{2}+4x+3-2x=7$$

Combine 4x and -2x to get 2x.

$$x^{2}+2x+3=7$$

Subtract 7 from both sides.

$$x^{2}+2x+3-7=0$$

Subtract 7 from 3 to get -4.

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

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

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

Square 2.

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

Multiply -4 times -4.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

Add 4 to 16.

$$x=\frac{-2±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-2±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{2} when ± is plus. Add -2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{2}$$

Divide -2+2\sqrt{5} by 2.

$$x=\sqrt{5}-1$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -2.

$$x=\frac{-2\sqrt{5}-2}{2}$$

Divide -2-2\sqrt{5} by 2.

$$x=-\sqrt{5}-1$$

The equation is now solved.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Steps for Completing the Square

Use the distributive property to multiply x+3 by x+1 and combine like terms.

$$x^{2}+4x+3=2x+7$$

Subtract 2x from both sides.

$$x^{2}+4x+3-2x=7$$

Combine 4x and -2x to get 2x.

$$x^{2}+2x+3=7$$

Subtract 3 from both sides.

$$x^{2}+2x=7-3$$

Subtract 3 from 7 to get 4.

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

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.

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

Square 1.

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

Add 4 to 1.

$$x^{2}+2x+1=5$$

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}.

$$\left(x+1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Solve for x

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\sqrt{5}-1\approx -3.236067977$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+3 by x+1 and combine like terms.

$$x^{2}+4x+3=2x+7$$

Subtract 2x from both sides.

$$x^{2}+4x+3-2x=7$$

Combine 4x and -2x to get 2x.

$$x^{2}+2x+3=7$$

Subtract 7 from both sides.

$$x^{2}+2x+3-7=0$$

Subtract 7 from 3 to get -4.

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

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

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

Square 2.

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

Multiply -4 times -4.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

Add 4 to 16.

$$x=\frac{-2±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-2±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{2} when ± is plus. Add -2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{2}$$

Divide -2+2\sqrt{5} by 2.

$$x=\sqrt{5}-1$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -2.

$$x=\frac{-2\sqrt{5}-2}{2}$$

Divide -2-2\sqrt{5} by 2.

$$x=-\sqrt{5}-1$$

The equation is now solved.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Steps for Completing the Square

Use the distributive property to multiply x+3 by x+1 and combine like terms.

$$x^{2}+4x+3=2x+7$$

Subtract 2x from both sides.

$$x^{2}+4x+3-2x=7$$

Combine 4x and -2x to get 2x.

$$x^{2}+2x+3=7$$

Subtract 3 from both sides.

$$x^{2}+2x=7-3$$

Subtract 3 from 7 to get 4.

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

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.

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

Square 1.

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

Add 4 to 1.

$$x^{2}+2x+1=5$$

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}.

$$\left(x+1\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

Subtract 1 from both sides of the equation.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$