0=2t^2-9

Solution for 0=2t^2-9 equation:

0=2t^2-9

We move all terms to the left:

0-(2t^2-9)=0

We add all the numbers together, and all the variables

-(2t^2-9)=0

We get rid of parentheses

-2t^2+9=0

a = -2; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-2)·9
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*-2}=\frac{0-6\sqrt{2}}{-4}
=-\frac{6\sqrt{2}}{-4}
=-\frac{3\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*-2}=\frac{0+6\sqrt{2}}{-4}
=\frac{6\sqrt{2}}{-4}
=\frac{3\sqrt{2}}{-2}
$