Solve d(10d-9)+1=0

$$d ( 10 d - 9 ) + 1 = 0$$

Solve for d

$$d=\frac{\sqrt{41}+9}{20}\approx 0.770156212$$

$$d=\frac{9-\sqrt{41}}{20}\approx 0.129843788$$

Steps Using the Quadratic Formula

Use the distributive property to multiply d by 10d-9.

$$10d^{2}-9d+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -9 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 10}}{2\times 10}$$

Square -9.

$$d=\frac{-\left(-9\right)±\sqrt{81-4\times 10}}{2\times 10}$$

Multiply -4 times 10.

$$d=\frac{-\left(-9\right)±\sqrt{81-40}}{2\times 10}$$

Add 81 to -40.

$$d=\frac{-\left(-9\right)±\sqrt{41}}{2\times 10}$$

The opposite of -9 is 9.

$$d=\frac{9±\sqrt{41}}{2\times 10}$$

Multiply 2 times 10.

$$d=\frac{9±\sqrt{41}}{20}$$

Now solve the equation d=\frac{9±\sqrt{41}}{20} when ± is plus. Add 9 to \sqrt{41}.

$$d=\frac{\sqrt{41}+9}{20}$$

Now solve the equation d=\frac{9±\sqrt{41}}{20} when ± is minus. Subtract \sqrt{41} from 9.

$$d=\frac{9-\sqrt{41}}{20}$$

The equation is now solved.

$$d=\frac{\sqrt{41}+9}{20} d=\frac{9-\sqrt{41}}{20}$$

Steps for Completing the Square

Use the distributive property to multiply d by 10d-9.

$$10d^{2}-9d+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$10d^{2}-9d=-1$$

Divide both sides by 10.

$$\frac{10d^{2}-9d}{10}=-\frac{1}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$d^{2}-\frac{9}{10}d=-\frac{1}{10}$$

Divide -\frac{9}{10}, the coefficient of the x term, by 2 to get -\frac{9}{20}. Then add the square of -\frac{9}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$d^{2}-\frac{9}{10}d+\left(-\frac{9}{20}\right)^{2}=-\frac{1}{10}+\left(-\frac{9}{20}\right)^{2}$$

Square -\frac{9}{20} by squaring both the numerator and the denominator of the fraction.

$$d^{2}-\frac{9}{10}d+\frac{81}{400}=-\frac{1}{10}+\frac{81}{400}$$

Add -\frac{1}{10} to \frac{81}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$d^{2}-\frac{9}{10}d+\frac{81}{400}=\frac{41}{400}$$

Factor d^{2}-\frac{9}{10}d+\frac{81}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(d-\frac{9}{20}\right)^{2}=\frac{41}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d-\frac{9}{20}\right)^{2}}=\sqrt{\frac{41}{400}}$$

Simplify.

$$d-\frac{9}{20}=\frac{\sqrt{41}}{20} d-\frac{9}{20}=-\frac{\sqrt{41}}{20}$$

Add \frac{9}{20} to both sides of the equation.

$$d=\frac{\sqrt{41}+9}{20} d=\frac{9-\sqrt{41}}{20}$$