Solve (3x+6)^2=(3x)^2+(2x+6)^2

$$( 3 x + 6 ) ^ { 2 } = ( 3 x ) ^ { 2 } + ( 2 x + 6 ) ^ { 2 }$$

Solve for x

$$x=3$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+6\right)^{2}.

$$9x^{2}+36x+36=\left(3x\right)^{2}+\left(2x+6\right)^{2}$$

Expand \left(3x\right)^{2}.

$$9x^{2}+36x+36=3^{2}x^{2}+\left(2x+6\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}+36x+36=9x^{2}+\left(2x+6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+6\right)^{2}.

$$9x^{2}+36x+36=9x^{2}+4x^{2}+24x+36$$

Combine 9x^{2} and 4x^{2} to get 13x^{2}.

$$9x^{2}+36x+36=13x^{2}+24x+36$$

Subtract 13x^{2} from both sides.

$$9x^{2}+36x+36-13x^{2}=24x+36$$

Combine 9x^{2} and -13x^{2} to get -4x^{2}.

$$-4x^{2}+36x+36=24x+36$$

Subtract 24x from both sides.

$$-4x^{2}+36x+36-24x=36$$

Combine 36x and -24x to get 12x.

$$-4x^{2}+12x+36=36$$

Subtract 36 from both sides.

$$-4x^{2}+12x+36-36=0$$

Subtract 36 from 36 to get 0.

$$-4x^{2}+12x=0$$

Factor out x.

$$x\left(-4x+12\right)=0$$

To find equation solutions, solve x=0 and -4x+12=0.

$$x=0 x=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+6\right)^{2}.

$$9x^{2}+36x+36=\left(3x\right)^{2}+\left(2x+6\right)^{2}$$

Expand \left(3x\right)^{2}.

$$9x^{2}+36x+36=3^{2}x^{2}+\left(2x+6\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}+36x+36=9x^{2}+\left(2x+6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+6\right)^{2}.

$$9x^{2}+36x+36=9x^{2}+4x^{2}+24x+36$$

Combine 9x^{2} and 4x^{2} to get 13x^{2}.

$$9x^{2}+36x+36=13x^{2}+24x+36$$

Subtract 13x^{2} from both sides.

$$9x^{2}+36x+36-13x^{2}=24x+36$$

Combine 9x^{2} and -13x^{2} to get -4x^{2}.

$$-4x^{2}+36x+36=24x+36$$

Subtract 24x from both sides.

$$-4x^{2}+36x+36-24x=36$$

Combine 36x and -24x to get 12x.

$$-4x^{2}+12x+36=36$$

Subtract 36 from both sides.

$$-4x^{2}+12x+36-36=0$$

Subtract 36 from 36 to get 0.

$$-4x^{2}+12x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 12 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}}}{2\left(-4\right)}$$

Take the square root of 12^{2}.

$$x=\frac{-12±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-12±12}{-8}$$

Now solve the equation x=\frac{-12±12}{-8} when ± is plus. Add -12 to 12.

$$x=\frac{0}{-8}$$

Divide 0 by -8.

$$x=0$$

Now solve the equation x=\frac{-12±12}{-8} when ± is minus. Subtract 12 from -12.

$$x=-\frac{24}{-8}$$

Divide -24 by -8.

$$x=3$$

The equation is now solved.

$$x=0 x=3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+6\right)^{2}.

$$9x^{2}+36x+36=\left(3x\right)^{2}+\left(2x+6\right)^{2}$$

Expand \left(3x\right)^{2}.

$$9x^{2}+36x+36=3^{2}x^{2}+\left(2x+6\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}+36x+36=9x^{2}+\left(2x+6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+6\right)^{2}.

$$9x^{2}+36x+36=9x^{2}+4x^{2}+24x+36$$

Combine 9x^{2} and 4x^{2} to get 13x^{2}.

$$9x^{2}+36x+36=13x^{2}+24x+36$$

Subtract 13x^{2} from both sides.

$$9x^{2}+36x+36-13x^{2}=24x+36$$

Combine 9x^{2} and -13x^{2} to get -4x^{2}.

$$-4x^{2}+36x+36=24x+36$$

Subtract 24x from both sides.

$$-4x^{2}+36x+36-24x=36$$

Combine 36x and -24x to get 12x.

$$-4x^{2}+12x+36=36$$

Subtract 36 from both sides.

$$-4x^{2}+12x=36-36$$

Subtract 36 from 36 to get 0.

$$-4x^{2}+12x=0$$

Divide both sides by -4.

$$\frac{-4x^{2}+12x}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{12}{-4}x=\frac{0}{-4}$$

Divide 12 by -4.

$$x^{2}-3x=\frac{0}{-4}$$

Divide 0 by -4.

$$x^{2}-3x=0$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{9}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3}{2} x-\frac{3}{2}=-\frac{3}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$