Solve y^2={left(2sqrt{3}right)}^2*2

$${ y }^{ 2 } = { \left(2 \sqrt{ 3 } \right) }^{ 2 } \times 2$$

Solve for y

$$y=2\sqrt{6}\approx 4.898979486$$

$$y=-2\sqrt{6}\approx -4.898979486$$

Steps by Finding Square Root

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

$$y^{2}=2\times 2^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$y^{2}=2\times 4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$y^{2}=2\times 4\times 3$$

Multiply 4 and 3 to get 12.

$$y^{2}=2\times 12$$

Multiply 2 and 12 to get 24.

$$y^{2}=24$$

Take the square root of both sides of the equation.

$$y=2\sqrt{6} y=-2\sqrt{6}$$

Steps Using the Quadratic Formula

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

$$y^{2}=2\times 2^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$y^{2}=2\times 4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$y^{2}=2\times 4\times 3$$

Multiply 4 and 3 to get 12.

$$y^{2}=2\times 12$$

Multiply 2 and 12 to get 24.

$$y^{2}=24$$

Subtract 24 from both sides.

$$y^{2}-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\left(-24\right)}}{2}$$

Square 0.

$$y=\frac{0±\sqrt{-4\left(-24\right)}}{2}$$

Multiply -4 times -24.

$$y=\frac{0±\sqrt{96}}{2}$$

Take the square root of 96.

$$y=\frac{0±4\sqrt{6}}{2}$$

Now solve the equation y=\frac{0±4\sqrt{6}}{2} when ± is plus.

$$y=2\sqrt{6}$$

Now solve the equation y=\frac{0±4\sqrt{6}}{2} when ± is minus.

$$y=-2\sqrt{6}$$

The equation is now solved.

$$y=2\sqrt{6} y=-2\sqrt{6}$$