9=t^2/3

Solution for 9=t^2/3 equation:

9=t^2/3

We move all terms to the left:

9-(t^2/3)=0

We get rid of parentheses

-t^2/3+9=0

We multiply all the terms by the denominator


-t^2+9*3=0

We add all the numbers together, and all the variables

-1t^2+27=0

a = -1; b = 0; c = +27;
Δ = b2-4ac
Δ = 02-4·(-1)·27
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*-1}=\frac{0-6\sqrt{3}}{-2}
=-\frac{6\sqrt{3}}{-2}
=-\frac{3\sqrt{3}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*-1}=\frac{0+6\sqrt{3}}{-2}
=\frac{6\sqrt{3}}{-2}
=\frac{3\sqrt{3}}{-1}
$