Solve (2+2d)^2=(2+d)(3+3d)

$$( 2 + 2 d ) ^ { 2 } = ( 2 + d ) ( 3 + 3 d )$$

Solve for d

$$d=-1$$

$$d=2$$

Steps Using Factoring

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

$$4+8d+4d^{2}=\left(2+d\right)\left(3+3d\right)$$

Use the distributive property to multiply 2+d by 3+3d and combine like terms.

$$4+8d+4d^{2}=6+9d+3d^{2}$$

Subtract 6 from both sides.

$$4+8d+4d^{2}-6=9d+3d^{2}$$

Subtract 6 from 4 to get -2.

$$-2+8d+4d^{2}=9d+3d^{2}$$

Subtract 9d from both sides.

$$-2+8d+4d^{2}-9d=3d^{2}$$

Combine 8d and -9d to get -d.

$$-2-d+4d^{2}=3d^{2}$$

Subtract 3d^{2} from both sides.

$$-2-d+4d^{2}-3d^{2}=0$$

Combine 4d^{2} and -3d^{2} to get d^{2}.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

$$a+b=-1 ab=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

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

$$\left(d-2\right)\left(d+1\right)$$

To find equation solutions, solve d-2=0 and d+1=0.

$$d=2 d=-1$$

Steps Using Factoring By Grouping

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

$$4+8d+4d^{2}=\left(2+d\right)\left(3+3d\right)$$

Use the distributive property to multiply 2+d by 3+3d and combine like terms.

$$4+8d+4d^{2}=6+9d+3d^{2}$$

Subtract 6 from both sides.

$$4+8d+4d^{2}-6=9d+3d^{2}$$

Subtract 6 from 4 to get -2.

$$-2+8d+4d^{2}=9d+3d^{2}$$

Subtract 9d from both sides.

$$-2+8d+4d^{2}-9d=3d^{2}$$

Combine 8d and -9d to get -d.

$$-2-d+4d^{2}=3d^{2}$$

Subtract 3d^{2} from both sides.

$$-2-d+4d^{2}-3d^{2}=0$$

Combine 4d^{2} and -3d^{2} to get d^{2}.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

$$a+b=-1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-2 b=1$$

Rewrite d^{2}-d-2 as \left(d^{2}-2d\right)+\left(d-2\right).

$$\left(d^{2}-2d\right)+\left(d-2\right)$$

Factor out d in d^{2}-2d.

$$d\left(d-2\right)+d-2$$

Factor out common term d-2 by using distributive property.

$$\left(d-2\right)\left(d+1\right)$$

To find equation solutions, solve d-2=0 and d+1=0.

$$d=2 d=-1$$

Steps Using the Quadratic Formula

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

$$4+8d+4d^{2}=\left(2+d\right)\left(3+3d\right)$$

Use the distributive property to multiply 2+d by 3+3d and combine like terms.

$$4+8d+4d^{2}=6+9d+3d^{2}$$

Subtract 6 from both sides.

$$4+8d+4d^{2}-6=9d+3d^{2}$$

Subtract 6 from 4 to get -2.

$$-2+8d+4d^{2}=9d+3d^{2}$$

Subtract 9d from both sides.

$$-2+8d+4d^{2}-9d=3d^{2}$$

Combine 8d and -9d to get -d.

$$-2-d+4d^{2}=3d^{2}$$

Subtract 3d^{2} from both sides.

$$-2-d+4d^{2}-3d^{2}=0$$

Combine 4d^{2} and -3d^{2} to get d^{2}.

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

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.

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

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

$$d=\frac{-\left(-1\right)±\sqrt{1-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$d=\frac{-\left(-1\right)±\sqrt{1+8}}{2}$$

Add 1 to 8.

$$d=\frac{-\left(-1\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$d=\frac{-\left(-1\right)±3}{2}$$

The opposite of -1 is 1.

$$d=\frac{1±3}{2}$$

Now solve the equation d=\frac{1±3}{2} when ± is plus. Add 1 to 3.

$$d=\frac{4}{2}$$

Divide 4 by 2.

$$d=2$$

Now solve the equation d=\frac{1±3}{2} when ± is minus. Subtract 3 from 1.

$$d=-\frac{2}{2}$$

Divide -2 by 2.

$$d=-1$$

The equation is now solved.

$$d=2 d=-1$$

Steps for Completing the Square

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

$$4+8d+4d^{2}=\left(2+d\right)\left(3+3d\right)$$

Use the distributive property to multiply 2+d by 3+3d and combine like terms.

$$4+8d+4d^{2}=6+9d+3d^{2}$$

Subtract 9d from both sides.

$$4+8d+4d^{2}-9d=6+3d^{2}$$

Combine 8d and -9d to get -d.

$$4-d+4d^{2}=6+3d^{2}$$

Subtract 3d^{2} from both sides.

$$4-d+4d^{2}-3d^{2}=6$$

Combine 4d^{2} and -3d^{2} to get d^{2}.

$$4-d+d^{2}=6$$

Subtract 4 from both sides.

$$-d+d^{2}=6-4$$

Subtract 4 from 6 to get 2.

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

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.

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

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.

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

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

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

Add 2 to \frac{1}{4}.

$$d^{2}-d+\frac{1}{4}=\frac{9}{4}$$

Factor d^{2}-d+\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}.

$$\left(d-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$d-\frac{1}{2}=\frac{3}{2} d-\frac{1}{2}=-\frac{3}{2}$$

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

$$d=2 d=-1$$