Solve T=2pisqrt{l/98}quad[2]

$$T = 2 \pi \sqrt { \frac { l } { 98 } } \quad [ 2 ]$$

Solve for l

$$l=\frac{49\times \left(\frac{T}{\pi }\right)^{2}}{8}$$

$$T\geq 0$$

Steps for Solving Radical Equations

Multiply 2 and 2 to get 4.

$$T=4\pi \sqrt{\frac{l}{98}}$$

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

$$4\pi \sqrt{\frac{l}{98}}=T$$

Divide both sides by 4\pi .

$$\frac{4\pi \sqrt{\frac{1}{98}l}}{4\pi }=\frac{T}{4\pi }$$

Dividing by 4\pi undoes the multiplication by 4\pi .

$$\sqrt{\frac{1}{98}l}=\frac{T}{4\pi }$$

Square both sides of the equation.

$$\frac{1}{98}l=\frac{T^{2}}{16\pi ^{2}}$$

Multiply both sides by 98.

$$\frac{\frac{1}{98}l}{\frac{1}{98}}=\frac{T^{2}}{\frac{1}{98}\times 16\pi ^{2}}$$

Dividing by \frac{1}{98} undoes the multiplication by \frac{1}{98}.

$$l=\frac{T^{2}}{\frac{1}{98}\times 16\pi ^{2}}$$

Divide \frac{T^{2}}{16\pi ^{2}} by \frac{1}{98} by multiplying \frac{T^{2}}{16\pi ^{2}} by the reciprocal of \frac{1}{98}.

$$l=\frac{49T^{2}}{8\pi ^{2}}$$

Solve for T

$$T=\frac{2\pi \sqrt{2l}}{7}$$

$$l\geq 0$$