Solve 2(a-1)^2-4=(a-1)^2

$$2 ( a - 1 ) ^ { 2 } - 4 = ( a - 1 ) ^ { 2 }$$

Solve for a

$$a=3$$

$$a=-1$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$2\left(a^{2}-2a+1\right)-4=\left(a-1\right)^{2}$$

Use the distributive property to multiply 2 by a^{2}-2a+1.

$$2a^{2}-4a+2-4=\left(a-1\right)^{2}$$

Subtract 4 from 2 to get -2.

$$2a^{2}-4a-2=\left(a-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$2a^{2}-4a-2=a^{2}-2a+1$$

Subtract a^{2} from both sides.

$$2a^{2}-4a-2-a^{2}=-2a+1$$

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

$$a^{2}-4a-2=-2a+1$$

Add 2a to both sides.

$$a^{2}-4a-2+2a=1$$

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

$$a^{2}-2a-2=1$$

Subtract 1 from both sides.

$$a^{2}-2a-2-1=0$$

Subtract 1 from -2 to get -3.

$$a^{2}-2a-3=0$$

To solve the equation, factor a^{2}-2a-3 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

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

$$\left(a-3\right)\left(a+1\right)$$

To find equation solutions, solve a-3=0 and a+1=0.

$$a=3 a=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$2\left(a^{2}-2a+1\right)-4=\left(a-1\right)^{2}$$

Use the distributive property to multiply 2 by a^{2}-2a+1.

$$2a^{2}-4a+2-4=\left(a-1\right)^{2}$$

Subtract 4 from 2 to get -2.

$$2a^{2}-4a-2=\left(a-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$2a^{2}-4a-2=a^{2}-2a+1$$

Subtract a^{2} from both sides.

$$2a^{2}-4a-2-a^{2}=-2a+1$$

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

$$a^{2}-4a-2=-2a+1$$

Add 2a to both sides.

$$a^{2}-4a-2+2a=1$$

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

$$a^{2}-2a-2=1$$

Subtract 1 from both sides.

$$a^{2}-2a-2-1=0$$

Subtract 1 from -2 to get -3.

$$a^{2}-2a-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite a^{2}-2a-3 as \left(a^{2}-3a\right)+\left(a-3\right).

$$\left(a^{2}-3a\right)+\left(a-3\right)$$

Factor out a in a^{2}-3a.

$$a\left(a-3\right)+a-3$$

Factor out common term a-3 by using distributive property.

$$\left(a-3\right)\left(a+1\right)$$

To find equation solutions, solve a-3=0 and a+1=0.

$$a=3 a=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$2\left(a^{2}-2a+1\right)-4=\left(a-1\right)^{2}$$

Use the distributive property to multiply 2 by a^{2}-2a+1.

$$2a^{2}-4a+2-4=\left(a-1\right)^{2}$$

Subtract 4 from 2 to get -2.

$$2a^{2}-4a-2=\left(a-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$2a^{2}-4a-2=a^{2}-2a+1$$

Subtract a^{2} from both sides.

$$2a^{2}-4a-2-a^{2}=-2a+1$$

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

$$a^{2}-4a-2=-2a+1$$

Add 2a to both sides.

$$a^{2}-4a-2+2a=1$$

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

$$a^{2}-2a-2=1$$

Subtract 1 from both sides.

$$a^{2}-2a-2-1=0$$

Subtract 1 from -2 to get -3.

$$a^{2}-2a-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)}}{2}$$

Square -2.

$$a=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$a=\frac{-\left(-2\right)±\sqrt{4+12}}{2}$$

Add 4 to 12.

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

Take the square root of 16.

$$a=\frac{-\left(-2\right)±4}{2}$$

The opposite of -2 is 2.

$$a=\frac{2±4}{2}$$

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

$$a=\frac{6}{2}$$

Divide 6 by 2.

$$a=3$$

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

$$a=-\frac{2}{2}$$

Divide -2 by 2.

$$a=-1$$

The equation is now solved.

$$a=3 a=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$2\left(a^{2}-2a+1\right)-4=\left(a-1\right)^{2}$$

Use the distributive property to multiply 2 by a^{2}-2a+1.

$$2a^{2}-4a+2-4=\left(a-1\right)^{2}$$

Subtract 4 from 2 to get -2.

$$2a^{2}-4a-2=\left(a-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$2a^{2}-4a-2=a^{2}-2a+1$$

Subtract a^{2} from both sides.

$$2a^{2}-4a-2-a^{2}=-2a+1$$

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

$$a^{2}-4a-2=-2a+1$$

Add 2a to both sides.

$$a^{2}-4a-2+2a=1$$

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

$$a^{2}-2a-2=1$$

Add 2 to both sides.

$$a^{2}-2a=1+2$$

Add 1 and 2 to get 3.

$$a^{2}-2a=3$$

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.

$$a^{2}-2a+1=3+1$$

Add 3 to 1.

$$a^{2}-2a+1=4$$

Factor a^{2}-2a+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}.

$$\left(a-1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-1\right)^{2}}=\sqrt{4}$$

Simplify.

$$a-1=2 a-1=-2$$

Add 1 to both sides of the equation.

$$a=3 a=-1$$