Solve (2y)^2=y^2+6^2

$$( 2 y ) ^ { 2 } = y ^ { 2 } + 6 ^ { 2 }$$

Solve for y

$$y=2\sqrt{3}\approx 3.464101615$$

$$y=-2\sqrt{3}\approx -3.464101615$$

Steps by Finding Square Root

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

$$2^{2}y^{2}=y^{2}+6^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4y^{2}=y^{2}+6^{2}$$

Calculate 6 to the power of 2 and get 36.

$$4y^{2}=y^{2}+36$$

Subtract y^{2} from both sides.

$$4y^{2}-y^{2}=36$$

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

$$3y^{2}=36$$

Divide both sides by 3.

$$y^{2}=\frac{36}{3}$$

Divide 36 by 3 to get 12.

$$y^{2}=12$$

Take the square root of both sides of the equation.

$$y=2\sqrt{3} y=-2\sqrt{3}$$

Steps Using the Quadratic Formula

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

$$2^{2}y^{2}=y^{2}+6^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4y^{2}=y^{2}+6^{2}$$

Calculate 6 to the power of 2 and get 36.

$$4y^{2}=y^{2}+36$$

Subtract y^{2} from both sides.

$$4y^{2}-y^{2}=36$$

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

$$3y^{2}=36$$

Subtract 36 from both sides.

$$3y^{2}-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times 3\left(-36\right)}}{2\times 3}$$

Square 0.

$$y=\frac{0±\sqrt{-4\times 3\left(-36\right)}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{0±\sqrt{-12\left(-36\right)}}{2\times 3}$$

Multiply -12 times -36.

$$y=\frac{0±\sqrt{432}}{2\times 3}$$

Take the square root of 432.

$$y=\frac{0±12\sqrt{3}}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{0±12\sqrt{3}}{6}$$

Now solve the equation y=\frac{0±12\sqrt{3}}{6} when ± is plus.

$$y=2\sqrt{3}$$

Now solve the equation y=\frac{0±12\sqrt{3}}{6} when ± is minus.

$$y=-2\sqrt{3}$$

The equation is now solved.

$$y=2\sqrt{3} y=-2\sqrt{3}$$