Solve sqrt{3L+t^2}=8-t

$$\sqrt{ 3L+ { t }^{ 2 } } = 8-t$$

Solve for L

$$L=\frac{64-16t}{3}$$

$$8-t\geq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$3L+t^{2}=\left(8-t\right)^{2}$$

Subtract t^{2} from both sides of the equation.

$$3L+t^{2}-t^{2}=\left(8-t\right)^{2}-t^{2}$$

Subtracting t^{2} from itself leaves 0.

$$3L=\left(8-t\right)^{2}-t^{2}$$

Subtract t^{2} from \left(8-t\right)^{2}.

$$3L=64-16t$$

Divide both sides by 3.

$$\frac{3L}{3}=\frac{64-16t}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$L=\frac{64-16t}{3}$$

Solve for L (complex solution)

$$L=\frac{64-16t}{3}$$

$$t=8\text{ or }arg(8-t)<\pi$$

Solve for t (complex solution)

$$t=-\frac{3L}{16}+4$$

$$L=-\frac{64}{3}\text{ or }arg(8-t)<\pi$$

Solve for t

$$t=-\frac{3L}{16}+4$$

$$L\geq -\frac{64}{3}$$