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

$$- 1 = \quad ( 2 ) ( x - 2 ) ( x + 2 ) - ( x + 1 ) ( x - 3 )$$

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 2 by x-2.

$$-1=\left(2x-4\right)\left(x+2\right)-\left(x+1\right)\left(x-3\right)$$

Use the distributive property to multiply 2x-4 by x+2 and combine like terms.

$$-1=2x^{2}-8-\left(x+1\right)\left(x-3\right)$$

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

$$-1=2x^{2}-8-\left(x^{2}-2x-3\right)$$

To find the opposite of x^{2}-2x-3, find the opposite of each term.

$$-1=2x^{2}-8-x^{2}+2x+3$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$-1=x^{2}-8+2x+3$$

Add -8 and 3 to get -5.

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

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

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

Add 1 to both sides.

$$x^{2}-5+2x+1=0$$

Add -5 and 1 to get -4.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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 2 by x-2.

$$-1=\left(2x-4\right)\left(x+2\right)-\left(x+1\right)\left(x-3\right)$$

Use the distributive property to multiply 2x-4 by x+2 and combine like terms.

$$-1=2x^{2}-8-\left(x+1\right)\left(x-3\right)$$

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

$$-1=2x^{2}-8-\left(x^{2}-2x-3\right)$$

To find the opposite of x^{2}-2x-3, find the opposite of each term.

$$-1=2x^{2}-8-x^{2}+2x+3$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$-1=x^{2}-8+2x+3$$

Add -8 and 3 to get -5.

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

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

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

Add 5 to both sides.

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

Add -1 and 5 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 2 by x-2.

$$-1=\left(2x-4\right)\left(x+2\right)-\left(x+1\right)\left(x-3\right)$$

Use the distributive property to multiply 2x-4 by x+2 and combine like terms.

$$-1=2x^{2}-8-\left(x+1\right)\left(x-3\right)$$

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

$$-1=2x^{2}-8-\left(x^{2}-2x-3\right)$$

To find the opposite of x^{2}-2x-3, find the opposite of each term.

$$-1=2x^{2}-8-x^{2}+2x+3$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$-1=x^{2}-8+2x+3$$

Add -8 and 3 to get -5.

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

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

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

Add 1 to both sides.

$$x^{2}-5+2x+1=0$$

Add -5 and 1 to get -4.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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 2 by x-2.

$$-1=\left(2x-4\right)\left(x+2\right)-\left(x+1\right)\left(x-3\right)$$

Use the distributive property to multiply 2x-4 by x+2 and combine like terms.

$$-1=2x^{2}-8-\left(x+1\right)\left(x-3\right)$$

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

$$-1=2x^{2}-8-\left(x^{2}-2x-3\right)$$

To find the opposite of x^{2}-2x-3, find the opposite of each term.

$$-1=2x^{2}-8-x^{2}+2x+3$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$-1=x^{2}-8+2x+3$$

Add -8 and 3 to get -5.

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

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

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

Add 5 to both sides.

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

Add -1 and 5 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.

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