3t/(t2-6) = √3

Basically I have to solve for t while incorporating the quadratic equation and I can't seem to get the correct answer. Any help would be appreciated. Thanks!

Originally Posted by OutOfMyMind
3t/(t2-6) = √3

Basically I have to solve for t while incorporating the quadratic equation and I can't seem to get the correct answer. Any help would be appreciated. Thanks!
$$\frac{3t}{t^2-t}=\sqrt{3}$$

$$\frac{\sqrt{3}t}{t^2-6}=1$$

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

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

you can use the quadratic formula here with

$$a=1, b=-\sqrt{3}, c=-6$$

or you can notice

$$t^2-\sqrt{3}t-6=(t-2\sqrt{3})(t+\sqrt{3})$$

and solve it that way.
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