0=-6t2+180

Solution for 0=-6t2+180 equation:

0=-6t^2+180

We move all terms to the left:

0-(-6t^2+180)=0

We add all the numbers together, and all the variables

-(-6t^2+180)=0

We get rid of parentheses

6t^2-180=0

a = 6; b = 0; c = -180;
Δ = b2-4ac
Δ = 02-4·6·(-180)
Δ = 4320
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:

$\sqrt{\Delta}=\sqrt{4320}=\sqrt{144*30}=\sqrt{144}*\sqrt{30}=12\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{30}}{2*6}=\frac{0-12\sqrt{30}}{12}
=-\frac{12\sqrt{30}}{12}
=-\sqrt{30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{30}}{2*6}=\frac{0+12\sqrt{30}}{12}
=\frac{12\sqrt{30}}{12}
=\sqrt{30}
$