Solve (10-b)(10+6b)=(10+2b)^2

$$( 10 - b ) ( 10 + 6 b ) = ( 10 + 2 b ) ^ { 2 }$$

Solve for b

$$b=1$$

$$b=0$$

Steps Using Factoring

Use the distributive property to multiply 10-b by 10+6b and combine like terms.

$$100+50b-6b^{2}=\left(10+2b\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10+2b\right)^{2}.

$$100+50b-6b^{2}=100+40b+4b^{2}$$

Subtract 100 from both sides.

$$100+50b-6b^{2}-100=40b+4b^{2}$$

Subtract 100 from 100 to get 0.

$$50b-6b^{2}=40b+4b^{2}$$

Subtract 40b from both sides.

$$50b-6b^{2}-40b=4b^{2}$$

Combine 50b and -40b to get 10b.

$$10b-6b^{2}=4b^{2}$$

Subtract 4b^{2} from both sides.

$$10b-6b^{2}-4b^{2}=0$$

Combine -6b^{2} and -4b^{2} to get -10b^{2}.

$$10b-10b^{2}=0$$

Factor out b.

$$b\left(10-10b\right)=0$$

To find equation solutions, solve b=0 and 10-10b=0.

$$b=0 b=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 10-b by 10+6b and combine like terms.

$$100+50b-6b^{2}=\left(10+2b\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10+2b\right)^{2}.

$$100+50b-6b^{2}=100+40b+4b^{2}$$

Subtract 100 from both sides.

$$100+50b-6b^{2}-100=40b+4b^{2}$$

Subtract 100 from 100 to get 0.

$$50b-6b^{2}=40b+4b^{2}$$

Subtract 40b from both sides.

$$50b-6b^{2}-40b=4b^{2}$$

Combine 50b and -40b to get 10b.

$$10b-6b^{2}=4b^{2}$$

Subtract 4b^{2} from both sides.

$$10b-6b^{2}-4b^{2}=0$$

Combine -6b^{2} and -4b^{2} to get -10b^{2}.

$$10b-10b^{2}=0$$

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of 10^{2}.

$$b=\frac{-10±10}{2\left(-10\right)}$$

Multiply 2 times -10.

$$b=\frac{-10±10}{-20}$$

Now solve the equation b=\frac{-10±10}{-20} when ± is plus. Add -10 to 10.

$$b=\frac{0}{-20}$$

Divide 0 by -20.

$$b=0$$

Now solve the equation b=\frac{-10±10}{-20} when ± is minus. Subtract 10 from -10.

$$b=-\frac{20}{-20}$$

Divide -20 by -20.

$$b=1$$

The equation is now solved.

$$b=0 b=1$$

Steps for Completing the Square

Use the distributive property to multiply 10-b by 10+6b and combine like terms.

$$100+50b-6b^{2}=\left(10+2b\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(10+2b\right)^{2}.

$$100+50b-6b^{2}=100+40b+4b^{2}$$

Subtract 40b from both sides.

$$100+50b-6b^{2}-40b=100+4b^{2}$$

Combine 50b and -40b to get 10b.

$$100+10b-6b^{2}=100+4b^{2}$$

Subtract 4b^{2} from both sides.

$$100+10b-6b^{2}-4b^{2}=100$$

Combine -6b^{2} and -4b^{2} to get -10b^{2}.

$$100+10b-10b^{2}=100$$

Subtract 100 from both sides.

$$10b-10b^{2}=100-100$$

Subtract 100 from 100 to get 0.

$$10b-10b^{2}=0$$

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.

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

Divide both sides by -10.

$$\frac{-10b^{2}+10b}{-10}=\frac{0}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$b^{2}+\frac{10}{-10}b=\frac{0}{-10}$$

Divide 10 by -10.

$$b^{2}-b=\frac{0}{-10}$$

Divide 0 by -10.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$b^{2}-b+\frac{1}{4}=\frac{1}{4}$$

Factor b^{2}-b+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$b-\frac{1}{2}=\frac{1}{2} b-\frac{1}{2}=-\frac{1}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$b=1 b=0$$