0=-16t^2+180t-420

Solution for 0=-16t^2+180t-420 equation:

0=-16t^2+180t-420

We move all terms to the left:

0-(-16t^2+180t-420)=0

We add all the numbers together, and all the variables

-(-16t^2+180t-420)=0

We get rid of parentheses

16t^2-180t+420=0

a = 16; b = -180; c = +420;
Δ = b2-4ac
Δ = -1802-4·16·420
Δ = 5520
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:

$\sqrt{\Delta}=\sqrt{5520}=\sqrt{16*345}=\sqrt{16}*\sqrt{345}=4\sqrt{345}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-180)-4\sqrt{345}}{2*16}=\frac{180-4\sqrt{345}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-180)+4\sqrt{345}}{2*16}=\frac{180+4\sqrt{345}}{32}
$