Solve -3=(x-2)^2-3

$$- 3 = ( x - 2 ) ^ { 2 } - 3$$

Solve for x

$$x=2$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Subtract 3 from 4 to get 1.

$$-3=x^{2}-4x+1$$

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

$$x^{2}-4x+1=-3$$

Add 3 to both sides.

$$x^{2}-4x+1+3=0$$

Add 1 and 3 to get 4.

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

To solve the equation, factor x^{2}-4x+4 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-4 ab=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-2\right)\left(x-2\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x-2=0.

$$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Subtract 3 from 4 to get 1.

$$-3=x^{2}-4x+1$$

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

$$x^{2}-4x+1=-3$$

Add 3 to both sides.

$$x^{2}-4x+1+3=0$$

Add 1 and 3 to get 4.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

Rewrite x^{2}-4x+4 as \left(x^{2}-2x\right)+\left(-2x+4\right).

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

Factor out x in the first and -2 in the second group.

$$x\left(x-2\right)-2\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(x-2\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x-2=0.

$$x=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Subtract 3 from 4 to get 1.

$$-3=x^{2}-4x+1$$

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

$$x^{2}-4x+1=-3$$

Add 3 to both sides.

$$x^{2}-4x+1+3=0$$

Add 1 and 3 to get 4.

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

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

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

Square -4.

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

Multiply -4 times 4.

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

Add 16 to -16.

$$x=\frac{-\left(-4\right)±\sqrt{0}}{2}$$

Take the square root of 0.

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

The opposite of -4 is 4.

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

Divide 4 by 2.

$$x=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

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

Subtract 3 from 4 to get 1.

$$-3=x^{2}-4x+1$$

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

$$x^{2}-4x+1=-3$$

Subtract 1 from both sides.

$$x^{2}-4x=-3-1$$

Subtract 1 from -3 to get -4.

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

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.

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

Square -2.

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

Add -4 to 4.

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

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}.

$$\left(x-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-2=0 x-2=0$$

Add 2 to both sides of the equation.

$$x=2 x=2$$

The equation is now solved. Solutions are the same.

$$x=2$$