Solve -2/tsqrt{t}=-sqrt{t}x+(t+2)sqrt{t}

$$- \frac { 2 } { t } \sqrt { t } = - \sqrt { t } x + ( t + 2 ) \sqrt { t }$$

Solve for x

$$x=t+2+\frac{2}{t}$$

$$t>0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by t.

$$\left(-\frac{2}{t}\right)\sqrt{t}t=\left(-\sqrt{t}\right)xt+\left(t+2\right)\sqrt{t}t$$

Express \left(-\frac{2}{t}\right)\sqrt{t} as a single fraction.

$$\frac{-2\sqrt{t}}{t}t=\left(-\sqrt{t}\right)xt+\left(t+2\right)\sqrt{t}t$$

Express \frac{-2\sqrt{t}}{t}t as a single fraction.

$$\frac{-2\sqrt{t}t}{t}=\left(-\sqrt{t}\right)xt+\left(t+2\right)\sqrt{t}t$$

Cancel out t in both numerator and denominator.

$$-2\sqrt{t}=\left(-\sqrt{t}\right)xt+\left(t+2\right)\sqrt{t}t$$

Use the distributive property to multiply t+2 by \sqrt{t}.

$$-2\sqrt{t}=\left(-\sqrt{t}\right)xt+\left(t\sqrt{t}+2\sqrt{t}\right)t$$

Use the distributive property to multiply t\sqrt{t}+2\sqrt{t} by t.

$$-2\sqrt{t}=\left(-\sqrt{t}\right)xt+\sqrt{t}t^{2}+2\sqrt{t}t$$

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

$$\left(-\sqrt{t}\right)xt+\sqrt{t}t^{2}+2\sqrt{t}t=-2\sqrt{t}$$

Subtract \sqrt{t}t^{2} from both sides.

$$\left(-\sqrt{t}\right)xt+2\sqrt{t}t=-2\sqrt{t}-\sqrt{t}t^{2}$$

Subtract 2\sqrt{t}t from both sides.

$$\left(-\sqrt{t}\right)xt=-2\sqrt{t}-\sqrt{t}t^{2}-2\sqrt{t}t$$

Multiply -1 and 2 to get -2.

$$-\sqrt{t}xt=-2\sqrt{t}-\sqrt{t}t^{2}-2\sqrt{t}t$$

The equation is in standard form.

$$\left(-\sqrt{t}t\right)x=-\sqrt{t}t^{2}-2\sqrt{t}t-2\sqrt{t}$$

Divide both sides by -\sqrt{t}t.

$$\frac{\left(-\sqrt{t}t\right)x}{-\sqrt{t}t}=-\frac{\sqrt{t}\left(t^{2}+2t+2\right)}{-\sqrt{t}t}$$

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

$$x=-\frac{\sqrt{t}\left(t^{2}+2t+2\right)}{-\sqrt{t}t}$$

Divide -\left(2+t^{2}+2t\right)\sqrt{t} by -\sqrt{t}t.

$$x=t+2+\frac{2}{t}$$

Solve for t

$$t=\frac{\sqrt{x^{2}-4x-4}+x-2}{2}$$

$$t=\frac{-\sqrt{x^{2}-4x-4}+x-2}{2}\text{, }x\geq 2\sqrt{2}+2$$