Solve b^2-20b+106=10

$${ b }^{ 2 } -20b+106 = 10$$

Solve for b

$$b=8$$

$$b=12$$

Steps Using Factoring

Subtract 10 from both sides.

$$b^{2}-20b+106-10=0$$

Subtract 10 from 106 to get 96.

$$b^{2}-20b+96=0$$

To solve the equation, factor b^{2}-20b+96 using formula b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). To find a and b, set up a system to be solved.

$$a+b=-20 ab=96$$

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 96.

$$-1,-96 -2,-48 -3,-32 -4,-24 -6,-16 -8,-12$$

Calculate the sum for each pair.

$$-1-96=-97 -2-48=-50 -3-32=-35 -4-24=-28 -6-16=-22 -8-12=-20$$

The solution is the pair that gives sum -20.

$$a=-12 b=-8$$

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

$$\left(b-12\right)\left(b-8\right)$$

To find equation solutions, solve b-12=0 and b-8=0.

$$b=12 b=8$$

Steps Using Factoring By Grouping

Subtract 10 from both sides.

$$b^{2}-20b+106-10=0$$

Subtract 10 from 106 to get 96.

$$b^{2}-20b+96=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb+96. To find a and b, set up a system to be solved.

$$a+b=-20 ab=1\times 96=96$$

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 96.

$$-1,-96 -2,-48 -3,-32 -4,-24 -6,-16 -8,-12$$

Calculate the sum for each pair.

$$-1-96=-97 -2-48=-50 -3-32=-35 -4-24=-28 -6-16=-22 -8-12=-20$$

The solution is the pair that gives sum -20.

$$a=-12 b=-8$$

Rewrite b^{2}-20b+96 as \left(b^{2}-12b\right)+\left(-8b+96\right).

$$\left(b^{2}-12b\right)+\left(-8b+96\right)$$

Factor out b in the first and -8 in the second group.

$$b\left(b-12\right)-8\left(b-12\right)$$

Factor out common term b-12 by using distributive property.

$$\left(b-12\right)\left(b-8\right)$$

To find equation solutions, solve b-12=0 and b-8=0.

$$b=12 b=8$$

Steps Using the Quadratic Formula

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.

$$b^{2}-20b+106=10$$

Subtract 10 from both sides of the equation.

$$b^{2}-20b+106-10=10-10$$

Subtracting 10 from itself leaves 0.

$$b^{2}-20b+106-10=0$$

Subtract 10 from 106.

$$b^{2}-20b+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -20 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 96}}{2}$$

Square -20.

$$b=\frac{-\left(-20\right)±\sqrt{400-4\times 96}}{2}$$

Multiply -4 times 96.

$$b=\frac{-\left(-20\right)±\sqrt{400-384}}{2}$$

Add 400 to -384.

$$b=\frac{-\left(-20\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$b=\frac{-\left(-20\right)±4}{2}$$

The opposite of -20 is 20.

$$b=\frac{20±4}{2}$$

Now solve the equation b=\frac{20±4}{2} when ± is plus. Add 20 to 4.

$$b=\frac{24}{2}$$

Divide 24 by 2.

$$b=12$$

Now solve the equation b=\frac{20±4}{2} when ± is minus. Subtract 4 from 20.

$$b=\frac{16}{2}$$

Divide 16 by 2.

$$b=8$$

The equation is now solved.

$$b=12 b=8$$

Steps for Completing the Square

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.

$$b^{2}-20b+106=10$$

Subtract 106 from both sides of the equation.

$$b^{2}-20b+106-106=10-106$$

Subtracting 106 from itself leaves 0.

$$b^{2}-20b=10-106$$

Subtract 106 from 10.

$$b^{2}-20b=-96$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-20b+\left(-10\right)^{2}=-96+\left(-10\right)^{2}$$

Square -10.

$$b^{2}-20b+100=-96+100$$

Add -96 to 100.

$$b^{2}-20b+100=4$$

Factor b^{2}-20b+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-10\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-10\right)^{2}}=\sqrt{4}$$

Simplify.

$$b-10=2 b-10=-2$$

Add 10 to both sides of the equation.

$$b=12 b=8$$