Solve (3x)^2=(2x)^2+(2sqrt{5})^2

$$( 3 x ) ^ { 2 } = ( 2 x ) ^ { 2 } + ( 2 \sqrt { 5 } ) ^ { 2 }$$

Solve for x

$$x=-2$$

$$x=2$$

Steps Using Difference of Squares

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

$$3^{2}x^{2}=\left(2x\right)^{2}+\left(2\sqrt{5}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}=\left(2x\right)^{2}+\left(2\sqrt{5}\right)^{2}$$

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

$$9x^{2}=2^{2}x^{2}+\left(2\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9x^{2}=4x^{2}+\left(2\sqrt{5}\right)^{2}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$9x^{2}=4x^{2}+2^{2}\left(\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9x^{2}=4x^{2}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9x^{2}=4x^{2}+4\times 5$$

Multiply 4 and 5 to get 20.

$$9x^{2}=4x^{2}+20$$

Subtract 4x^{2} from both sides.

$$9x^{2}-4x^{2}=20$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}=20$$

Subtract 20 from both sides.

$$5x^{2}-20=0$$

Divide both sides by 5.

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

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

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

$$x=2 x=-2$$

Steps by Finding Square Root

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

$$3^{2}x^{2}=\left(2x\right)^{2}+\left(2\sqrt{5}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}=\left(2x\right)^{2}+\left(2\sqrt{5}\right)^{2}$$

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

$$9x^{2}=2^{2}x^{2}+\left(2\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9x^{2}=4x^{2}+\left(2\sqrt{5}\right)^{2}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$9x^{2}=4x^{2}+2^{2}\left(\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9x^{2}=4x^{2}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9x^{2}=4x^{2}+4\times 5$$

Multiply 4 and 5 to get 20.

$$9x^{2}=4x^{2}+20$$

Subtract 4x^{2} from both sides.

$$9x^{2}-4x^{2}=20$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}=20$$

Divide both sides by 5.

$$x^{2}=\frac{20}{5}$$

Divide 20 by 5 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

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

$$3^{2}x^{2}=\left(2x\right)^{2}+\left(2\sqrt{5}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}=\left(2x\right)^{2}+\left(2\sqrt{5}\right)^{2}$$

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

$$9x^{2}=2^{2}x^{2}+\left(2\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9x^{2}=4x^{2}+\left(2\sqrt{5}\right)^{2}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$9x^{2}=4x^{2}+2^{2}\left(\sqrt{5}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$9x^{2}=4x^{2}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9x^{2}=4x^{2}+4\times 5$$

Multiply 4 and 5 to get 20.

$$9x^{2}=4x^{2}+20$$

Subtract 4x^{2} from both sides.

$$9x^{2}-4x^{2}=20$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}=20$$

Subtract 20 from both sides.

$$5x^{2}-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 0 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 5\left(-20\right)}}{2\times 5}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5\left(-20\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{0±\sqrt{-20\left(-20\right)}}{2\times 5}$$

Multiply -20 times -20.

$$x=\frac{0±\sqrt{400}}{2\times 5}$$

Take the square root of 400.

$$x=\frac{0±20}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{0±20}{10}$$

Now solve the equation x=\frac{0±20}{10} when ± is plus. Divide 20 by 10.

$$x=2$$

Now solve the equation x=\frac{0±20}{10} when ± is minus. Divide -20 by 10.

$$x=-2$$

The equation is now solved.

$$x=2 x=-2$$