Solve sqrt{128+sqrt{260-sqrt{16}}}=text{feedt}?

$$\sqrt { 128 + \sqrt { 260 - \sqrt { 16 } } } = \text { feedt } ?$$

Solve for d

$$d=\frac{12}{e^{2}ft}$$

$$t\neq 0\text{ and }f\neq 0$$

Steps for Solving Linear Equation

Multiply e and e to get e^{2}.

$$\sqrt{128+\sqrt{260-\sqrt{16}}}=fe^{2}dt$$

Calculate the square root of 16 and get 4.

$$\sqrt{128+\sqrt{260-4}}=fe^{2}dt$$

Subtract 4 from 260 to get 256.

$$\sqrt{128+\sqrt{256}}=fe^{2}dt$$

Calculate the square root of 256 and get 16.

$$\sqrt{128+16}=fe^{2}dt$$

Add 128 and 16 to get 144.

$$\sqrt{144}=fe^{2}dt$$

Calculate the square root of 144 and get 12.

$$12=fe^{2}dt$$

Swap sides so that all variable terms are on the left hand side.

$$fe^{2}dt=12$$

The equation is in standard form.

$$e^{2}ftd=12$$

Divide both sides by fe^{2}t.

$$\frac{e^{2}ftd}{e^{2}ft}=\frac{12}{e^{2}ft}$$

Dividing by fe^{2}t undoes the multiplication by fe^{2}t.

$$d=\frac{12}{e^{2}ft}$$

Solve for f

$$f=\frac{12}{e^{2}dt}$$

$$t\neq 0\text{ and }d\neq 0$$

Steps for Solving Linear Equation

Multiply e and e to get e^{2}.

$$\sqrt{128+\sqrt{260-\sqrt{16}}}=fe^{2}dt$$

Calculate the square root of 16 and get 4.

$$\sqrt{128+\sqrt{260-4}}=fe^{2}dt$$

Subtract 4 from 260 to get 256.

$$\sqrt{128+\sqrt{256}}=fe^{2}dt$$

Calculate the square root of 256 and get 16.

$$\sqrt{128+16}=fe^{2}dt$$

Add 128 and 16 to get 144.

$$\sqrt{144}=fe^{2}dt$$

Calculate the square root of 144 and get 12.

$$12=fe^{2}dt$$

Swap sides so that all variable terms are on the left hand side.

$$fe^{2}dt=12$$

The equation is in standard form.

$$e^{2}dtf=12$$

Divide both sides by e^{2}dt.

$$\frac{e^{2}dtf}{e^{2}dt}=\frac{12}{e^{2}dt}$$

Dividing by e^{2}dt undoes the multiplication by e^{2}dt.

$$f=\frac{12}{e^{2}dt}$$