Solve (1-10a)(11-10a)=5

$$( 1 - 10 a ) ( 11 - 10 a ) = 5$$

Solve for a

$$a=-\frac{\sqrt{30}}{10}+\frac{3}{5}\approx 0.052277442$$

$$a=\frac{\sqrt{30}}{10}+\frac{3}{5}\approx 1.147722558$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1-10a by 11-10a and combine like terms.

$$11-120a+100a^{2}=5$$

Subtract 5 from both sides.

$$11-120a+100a^{2}-5=0$$

Subtract 5 from 11 to get 6.

$$6-120a+100a^{2}=0$$

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.

$$100a^{2}-120a+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 100 for a, -120 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-120\right)±\sqrt{\left(-120\right)^{2}-4\times 100\times 6}}{2\times 100}$$

Square -120.

$$a=\frac{-\left(-120\right)±\sqrt{14400-4\times 100\times 6}}{2\times 100}$$

Multiply -4 times 100.

$$a=\frac{-\left(-120\right)±\sqrt{14400-400\times 6}}{2\times 100}$$

Multiply -400 times 6.

$$a=\frac{-\left(-120\right)±\sqrt{14400-2400}}{2\times 100}$$

Add 14400 to -2400.

$$a=\frac{-\left(-120\right)±\sqrt{12000}}{2\times 100}$$

Take the square root of 12000.

$$a=\frac{-\left(-120\right)±20\sqrt{30}}{2\times 100}$$

The opposite of -120 is 120.

$$a=\frac{120±20\sqrt{30}}{2\times 100}$$

Multiply 2 times 100.

$$a=\frac{120±20\sqrt{30}}{200}$$

Now solve the equation a=\frac{120±20\sqrt{30}}{200} when ± is plus. Add 120 to 20\sqrt{30}.

$$a=\frac{20\sqrt{30}+120}{200}$$

Divide 120+20\sqrt{30} by 200.

$$a=\frac{\sqrt{30}}{10}+\frac{3}{5}$$

Now solve the equation a=\frac{120±20\sqrt{30}}{200} when ± is minus. Subtract 20\sqrt{30} from 120.

$$a=\frac{120-20\sqrt{30}}{200}$$

Divide 120-20\sqrt{30} by 200.

$$a=-\frac{\sqrt{30}}{10}+\frac{3}{5}$$

The equation is now solved.

$$a=\frac{\sqrt{30}}{10}+\frac{3}{5} a=-\frac{\sqrt{30}}{10}+\frac{3}{5}$$

Steps for Completing the Square

Use the distributive property to multiply 1-10a by 11-10a and combine like terms.

$$11-120a+100a^{2}=5$$

Subtract 11 from both sides.

$$-120a+100a^{2}=5-11$$

Subtract 11 from 5 to get -6.

$$-120a+100a^{2}=-6$$

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.

$$100a^{2}-120a=-6$$

Divide both sides by 100.

$$\frac{100a^{2}-120a}{100}=-\frac{6}{100}$$

Dividing by 100 undoes the multiplication by 100.

$$a^{2}+\left(-\frac{120}{100}\right)a=-\frac{6}{100}$$

Reduce the fraction \frac{-120}{100} to lowest terms by extracting and canceling out 20.

$$a^{2}-\frac{6}{5}a=-\frac{6}{100}$$

Reduce the fraction \frac{-6}{100} to lowest terms by extracting and canceling out 2.

$$a^{2}-\frac{6}{5}a=-\frac{3}{50}$$

Divide -\frac{6}{5}, the coefficient of the x term, by 2 to get -\frac{3}{5}. Then add the square of -\frac{3}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{6}{5}a+\left(-\frac{3}{5}\right)^{2}=-\frac{3}{50}+\left(-\frac{3}{5}\right)^{2}$$

Square -\frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{6}{5}a+\frac{9}{25}=-\frac{3}{50}+\frac{9}{25}$$

Add -\frac{3}{50} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{6}{5}a+\frac{9}{25}=\frac{3}{10}$$

Factor a^{2}-\frac{6}{5}a+\frac{9}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{3}{5}\right)^{2}=\frac{3}{10}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{3}{5}\right)^{2}}=\sqrt{\frac{3}{10}}$$

Simplify.

$$a-\frac{3}{5}=\frac{\sqrt{30}}{10} a-\frac{3}{5}=-\frac{\sqrt{30}}{10}$$

Add \frac{3}{5} to both sides of the equation.

$$a=\frac{\sqrt{30}}{10}+\frac{3}{5} a=-\frac{\sqrt{30}}{10}+\frac{3}{5}$$