Solve frac{y}2^2+8=16

$$\frac { y } { 2 } ^ { 2 } + 8 = 16$$

Solve for y

$$y=4\sqrt{2}\approx 5.656854249$$

$$y=-4\sqrt{2}\approx -5.656854249$$

Steps by Finding Square Root

To raise \frac{y}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{y^{2}}{2^{2}}+8=16$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 8 times \frac{2^{2}}{2^{2}}.

$$\frac{y^{2}}{2^{2}}+\frac{8\times 2^{2}}{2^{2}}=16$$

Since \frac{y^{2}}{2^{2}} and \frac{8\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{y^{2}+8\times 2^{2}}{2^{2}}=16$$

Do the multiplications in y^{2}+8\times 2^{2}.

$$\frac{y^{2}+32}{2^{2}}=16$$

Calculate 2 to the power of 2 and get 4.

$$\frac{y^{2}+32}{4}=16$$

Divide each term of y^{2}+32 by 4 to get \frac{1}{4}y^{2}+8.

$$\frac{1}{4}y^{2}+8=16$$

Subtract 8 from both sides.

$$\frac{1}{4}y^{2}=16-8$$

Subtract 8 from 16 to get 8.

$$\frac{1}{4}y^{2}=8$$

Multiply both sides by 4, the reciprocal of \frac{1}{4}.

$$y^{2}=8\times 4$$

Multiply 8 and 4 to get 32.

$$y^{2}=32$$

Take the square root of both sides of the equation.

$$y=4\sqrt{2} y=-4\sqrt{2}$$