Solve (1-h)^2+1=5

$$( 1 - h ) ^ { 2 } + 1 = 5$$

Solve for h

$$h=-1$$

$$h=3$$

Steps Using Factoring

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

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

Add 1 and 1 to get 2.

$$2-2h+h^{2}=5$$

Subtract 5 from both sides.

$$2-2h+h^{2}-5=0$$

Subtract 5 from 2 to get -3.

$$-3-2h+h^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor h^{2}-2h-3 using formula h^{2}+\left(a+b\right)h+ab=\left(h+a\right)\left(h+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(h+a\right)\left(h+b\right) using the obtained values.

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

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

$$h=3 h=-1$$

Steps Using Factoring By Grouping

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

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

Add 1 and 1 to get 2.

$$2-2h+h^{2}=5$$

Subtract 5 from both sides.

$$2-2h+h^{2}-5=0$$

Subtract 5 from 2 to get -3.

$$-3-2h+h^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as h^{2}+ah+bh-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 h^{2}-2h-3 as \left(h^{2}-3h\right)+\left(h-3\right).

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

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

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

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

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

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

$$h=3 h=-1$$

Steps Using the Quadratic Formula

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

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

Add 1 and 1 to get 2.

$$2-2h+h^{2}=5$$

Subtract 5 from both sides.

$$2-2h+h^{2}-5=0$$

Subtract 5 from 2 to get -3.

$$-3-2h+h^{2}=0$$

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.

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

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}.

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

Square -2.

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

Multiply -4 times -3.

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

Add 4 to 12.

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

Take the square root of 16.

$$h=\frac{-\left(-2\right)±4}{2}$$

The opposite of -2 is 2.

$$h=\frac{2±4}{2}$$

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

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

Divide 6 by 2.

$$h=3$$

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

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

Divide -2 by 2.

$$h=-1$$

The equation is now solved.

$$h=3 h=-1$$

Steps for Completing the Square

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

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

Add 1 and 1 to get 2.

$$2-2h+h^{2}=5$$

Subtract 2 from both sides.

$$-2h+h^{2}=5-2$$

Subtract 2 from 5 to get 3.

$$-2h+h^{2}=3$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

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.

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

Add 3 to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$h-1=2 h-1=-2$$

Add 1 to both sides of the equation.

$$h=3 h=-1$$