Solve (x+1)(x-3)=2x-5

$$( x + 1 ) ( x - 3 ) = 2 x - 5$$

Solve for x

$$x=\sqrt{2}+2\approx 3.414213562$$

$$x=2-\sqrt{2}\approx 0.585786438$$

Steps Using the Quadratic Formula

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

$$x^{2}-2x-3=2x-5$$

Subtract 2x from both sides.

$$x^{2}-2x-3-2x=-5$$

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

$$x^{2}-4x-3=-5$$

Add 5 to both sides.

$$x^{2}-4x-3+5=0$$

Add -3 and 5 to get 2.

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

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

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

Square -4.

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

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8}}{2}$$

Add 16 to -8.

$$x=\frac{-\left(-4\right)±\sqrt{8}}{2}$$

Take the square root of 8.

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

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{2}}{2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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

$$x^{2}-2x-3=2x-5$$

Subtract 2x from both sides.

$$x^{2}-2x-3-2x=-5$$

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

$$x^{2}-4x-3=-5$$

Add 3 to both sides.

$$x^{2}-4x=-5+3$$

Add -5 and 3 to get -2.

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

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.

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

Square -2.

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

Add -2 to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2 to both sides of the equation.

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