Solve y=sqrt{4(2sqrt{2}+2)-{left(2sqrt{2}+2right)}^2+5}

$$y= \sqrt{ 4(2 \sqrt{ 2 } +2)- { \left(2 \sqrt{ 2 } +2 \right) }^{ 2 } +5 }$$

Solve for y

$$y=1$$

Steps for Solving Linear Equation

Use the distributive property to multiply 4 by 2\sqrt{2}+2.

$$y=\sqrt{8\sqrt{2}+8-\left(2\sqrt{2}+2\right)^{2}+5}$$

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

$$y=\sqrt{8\sqrt{2}+8-\left(4\left(\sqrt{2}\right)^{2}+8\sqrt{2}+4\right)+5}$$

The square of \sqrt{2} is 2.

$$y=\sqrt{8\sqrt{2}+8-\left(4\times 2+8\sqrt{2}+4\right)+5}$$

Multiply 4 and 2 to get 8.

$$y=\sqrt{8\sqrt{2}+8-\left(8+8\sqrt{2}+4\right)+5}$$

Add 8 and 4 to get 12.

$$y=\sqrt{8\sqrt{2}+8-\left(12+8\sqrt{2}\right)+5}$$

To find the opposite of 12+8\sqrt{2}, find the opposite of each term.

$$y=\sqrt{8\sqrt{2}+8-12-8\sqrt{2}+5}$$

Subtract 12 from 8 to get -4.

$$y=\sqrt{8\sqrt{2}-4-8\sqrt{2}+5}$$

Combine 8\sqrt{2} and -8\sqrt{2} to get 0.

$$y=\sqrt{-4+5}$$

Add -4 and 5 to get 1.

$$y=\sqrt{1}$$

Calculate the square root of 1 and get 1.

$$y=1$$

Assign y

$$y≔1$$