Solve (5-d)(5+10d)=(5+2d)^2

$$( 5 - d ) ( 5 + 10 d ) = ( 5 + 2 d ) ^ { 2 }$$

Solve for d

$$d = \frac{25}{14} = 1\frac{11}{14} \approx 1.785714286$$

$$d=0$$

Steps Using Factoring

Use the distributive property to multiply 5-d by 5+10d and combine like terms.

$$25+45d-10d^{2}=\left(5+2d\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+2d\right)^{2}.

$$25+45d-10d^{2}=25+20d+4d^{2}$$

Subtract 25 from both sides.

$$25+45d-10d^{2}-25=20d+4d^{2}$$

Subtract 25 from 25 to get 0.

$$45d-10d^{2}=20d+4d^{2}$$

Subtract 20d from both sides.

$$45d-10d^{2}-20d=4d^{2}$$

Combine 45d and -20d to get 25d.

$$25d-10d^{2}=4d^{2}$$

Subtract 4d^{2} from both sides.

$$25d-10d^{2}-4d^{2}=0$$

Combine -10d^{2} and -4d^{2} to get -14d^{2}.

$$25d-14d^{2}=0$$

Factor out d.

$$d\left(25-14d\right)=0$$

To find equation solutions, solve d=0 and 25-14d=0.

$$d=0 d=\frac{25}{14}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5-d by 5+10d and combine like terms.

$$25+45d-10d^{2}=\left(5+2d\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+2d\right)^{2}.

$$25+45d-10d^{2}=25+20d+4d^{2}$$

Subtract 25 from both sides.

$$25+45d-10d^{2}-25=20d+4d^{2}$$

Subtract 25 from 25 to get 0.

$$45d-10d^{2}=20d+4d^{2}$$

Subtract 20d from both sides.

$$45d-10d^{2}-20d=4d^{2}$$

Combine 45d and -20d to get 25d.

$$25d-10d^{2}=4d^{2}$$

Subtract 4d^{2} from both sides.

$$25d-10d^{2}-4d^{2}=0$$

Combine -10d^{2} and -4d^{2} to get -14d^{2}.

$$25d-14d^{2}=0$$

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.

$$-14d^{2}+25d=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -14 for a, 25 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{-25±\sqrt{25^{2}}}{2\left(-14\right)}$$

Take the square root of 25^{2}.

$$d=\frac{-25±25}{2\left(-14\right)}$$

Multiply 2 times -14.

$$d=\frac{-25±25}{-28}$$

Now solve the equation d=\frac{-25±25}{-28} when ± is plus. Add -25 to 25.

$$d=\frac{0}{-28}$$

Divide 0 by -28.

$$d=0$$

Now solve the equation d=\frac{-25±25}{-28} when ± is minus. Subtract 25 from -25.

$$d=-\frac{50}{-28}$$

Reduce the fraction \frac{-50}{-28} to lowest terms by extracting and canceling out 2.

$$d=\frac{25}{14}$$

The equation is now solved.

$$d=0 d=\frac{25}{14}$$

Steps for Completing the Square

Use the distributive property to multiply 5-d by 5+10d and combine like terms.

$$25+45d-10d^{2}=\left(5+2d\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+2d\right)^{2}.

$$25+45d-10d^{2}=25+20d+4d^{2}$$

Subtract 20d from both sides.

$$25+45d-10d^{2}-20d=25+4d^{2}$$

Combine 45d and -20d to get 25d.

$$25+25d-10d^{2}=25+4d^{2}$$

Subtract 4d^{2} from both sides.

$$25+25d-10d^{2}-4d^{2}=25$$

Combine -10d^{2} and -4d^{2} to get -14d^{2}.

$$25+25d-14d^{2}=25$$

Subtract 25 from both sides.

$$25d-14d^{2}=25-25$$

Subtract 25 from 25 to get 0.

$$25d-14d^{2}=0$$

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.

$$-14d^{2}+25d=0$$

Divide both sides by -14.

$$\frac{-14d^{2}+25d}{-14}=\frac{0}{-14}$$

Dividing by -14 undoes the multiplication by -14.

$$d^{2}+\frac{25}{-14}d=\frac{0}{-14}$$

Divide 25 by -14.

$$d^{2}-\frac{25}{14}d=\frac{0}{-14}$$

Divide 0 by -14.

$$d^{2}-\frac{25}{14}d=0$$

Divide -\frac{25}{14}, the coefficient of the x term, by 2 to get -\frac{25}{28}. Then add the square of -\frac{25}{28} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$d^{2}-\frac{25}{14}d+\left(-\frac{25}{28}\right)^{2}=\left(-\frac{25}{28}\right)^{2}$$

Square -\frac{25}{28} by squaring both the numerator and the denominator of the fraction.

$$d^{2}-\frac{25}{14}d+\frac{625}{784}=\frac{625}{784}$$

Factor d^{2}-\frac{25}{14}d+\frac{625}{784}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(d-\frac{25}{28}\right)^{2}=\frac{625}{784}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d-\frac{25}{28}\right)^{2}}=\sqrt{\frac{625}{784}}$$

Simplify.

$$d-\frac{25}{28}=\frac{25}{28} d-\frac{25}{28}=-\frac{25}{28}$$

Add \frac{25}{28} to both sides of the equation.

$$d=\frac{25}{14} d=0$$