Solve (14-b)^2+b^2=100

$$( 14 - b ) ^ { 2 } + b ^ { 2 } = 100$$

Solve for b

$$b=6$$

$$b=8$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-b\right)^{2}.

$$196-28b+b^{2}+b^{2}=100$$

Combine b^{2} and b^{2} to get 2b^{2}.

$$196-28b+2b^{2}=100$$

Subtract 100 from both sides.

$$196-28b+2b^{2}-100=0$$

Subtract 100 from 196 to get 96.

$$96-28b+2b^{2}=0$$

Divide both sides by 2.

$$48-14b+b^{2}=0$$

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

$$b^{2}-14b+48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb+48. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 48=48$$

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 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -14.

$$a=-8 b=-6$$

Rewrite b^{2}-14b+48 as \left(b^{2}-8b\right)+\left(-6b+48\right).

$$\left(b^{2}-8b\right)+\left(-6b+48\right)$$

Factor out b in the first and -6 in the second group.

$$b\left(b-8\right)-6\left(b-8\right)$$

Factor out common term b-8 by using distributive property.

$$\left(b-8\right)\left(b-6\right)$$

To find equation solutions, solve b-8=0 and b-6=0.

$$b=8 b=6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-b\right)^{2}.

$$196-28b+b^{2}+b^{2}=100$$

Combine b^{2} and b^{2} to get 2b^{2}.

$$196-28b+2b^{2}=100$$

Subtract 100 from both sides.

$$196-28b+2b^{2}-100=0$$

Subtract 100 from 196 to get 96.

$$96-28b+2b^{2}=0$$

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.

$$2b^{2}-28b+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -28 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 2\times 96}}{2\times 2}$$

Square -28.

$$b=\frac{-\left(-28\right)±\sqrt{784-4\times 2\times 96}}{2\times 2}$$

Multiply -4 times 2.

$$b=\frac{-\left(-28\right)±\sqrt{784-8\times 96}}{2\times 2}$$

Multiply -8 times 96.

$$b=\frac{-\left(-28\right)±\sqrt{784-768}}{2\times 2}$$

Add 784 to -768.

$$b=\frac{-\left(-28\right)±\sqrt{16}}{2\times 2}$$

Take the square root of 16.

$$b=\frac{-\left(-28\right)±4}{2\times 2}$$

The opposite of -28 is 28.

$$b=\frac{28±4}{2\times 2}$$

Multiply 2 times 2.

$$b=\frac{28±4}{4}$$

Now solve the equation b=\frac{28±4}{4} when ± is plus. Add 28 to 4.

$$b=\frac{32}{4}$$

Divide 32 by 4.

$$b=8$$

Now solve the equation b=\frac{28±4}{4} when ± is minus. Subtract 4 from 28.

$$b=\frac{24}{4}$$

Divide 24 by 4.

$$b=6$$

The equation is now solved.

$$b=8 b=6$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(14-b\right)^{2}.

$$196-28b+b^{2}+b^{2}=100$$

Combine b^{2} and b^{2} to get 2b^{2}.

$$196-28b+2b^{2}=100$$

Subtract 196 from both sides.

$$-28b+2b^{2}=100-196$$

Subtract 196 from 100 to get -96.

$$-28b+2b^{2}=-96$$

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.

$$2b^{2}-28b=-96$$

Divide both sides by 2.

$$\frac{2b^{2}-28b}{2}=-\frac{96}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$b^{2}+\left(-\frac{28}{2}\right)b=-\frac{96}{2}$$

Divide -28 by 2.

$$b^{2}-14b=-\frac{96}{2}$$

Divide -96 by 2.

$$b^{2}-14b=-48$$

Divide -14, the coefficient of the x term, by 2 to get -7. Then add the square of -7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-14b+\left(-7\right)^{2}=-48+\left(-7\right)^{2}$$

Square -7.

$$b^{2}-14b+49=-48+49$$

Add -48 to 49.

$$b^{2}-14b+49=1$$

Factor b^{2}-14b+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-7\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-7\right)^{2}}=\sqrt{1}$$

Simplify.

$$b-7=1 b-7=-1$$

Add 7 to both sides of the equation.

$$b=8 b=6$$