Solve t^2-sqrt{3}c-6=0

$$t ^ { 2 } - \sqrt { 3 } c - 6 = 0$$

Solve for c

$$c=\frac{\sqrt{3}\left(t^{2}-6\right)}{3}$$

Steps for Solving Linear Equation

Add 6 to both sides. Anything plus zero gives itself.

$$t^{2}-\sqrt{3}c=6$$

Subtract t^{2} from both sides.

$$-\sqrt{3}c=6-t^{2}$$

The equation is in standard form.

$$\left(-\sqrt{3}\right)c=6-t^{2}$$

Divide both sides by -\sqrt{3}.

$$\frac{\left(-\sqrt{3}\right)c}{-\sqrt{3}}=\frac{6-t^{2}}{-\sqrt{3}}$$

Dividing by -\sqrt{3} undoes the multiplication by -\sqrt{3}.

$$c=\frac{6-t^{2}}{-\sqrt{3}}$$

Divide 6-t^{2} by -\sqrt{3}.

$$c=\frac{\sqrt{3}t^{2}}{3}-2\sqrt{3}$$

Solve for t

$$t=\sqrt{\sqrt{3}c+6}$$

$$t=-\sqrt{\sqrt{3}c+6}\text{, }c\geq -2\sqrt{3}$$