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

$$( x - 1 ) ^ { 2 } + 2 x - 3 = 2$$

Solve for x

$$x=-2$$

$$x=2$$

Steps Using Difference of Squares

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

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

Combine -2x and 2x to get 0.

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

Subtract 3 from 1 to get -2.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

To find equation solutions, solve x-2=0 and x+2=0.

$$x=2 x=-2$$

Steps by Finding Square Root

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

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

Combine -2x and 2x to get 0.

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

Subtract 3 from 1 to get -2.

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

Add 2 to both sides.

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

Add 2 and 2 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

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

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

Combine -2x and 2x to get 0.

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

Subtract 3 from 1 to get -2.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

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

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

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

Square 0.

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

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{0±4}{2}$$

Now solve the equation x=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=2 x=-2$$