0=-16t^2+180t+3

Solution for 0=-16t^2+180t+3 equation:

0=-16t^2+180t+3

We move all terms to the left:

0-(-16t^2+180t+3)=0

We add all the numbers together, and all the variables

-(-16t^2+180t+3)=0

We get rid of parentheses

16t^2-180t-3=0

a = 16; b = -180; c = -3;
Δ = b2-4ac
Δ = -1802-4·16·(-3)
Δ = 32592
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:

$\sqrt{\Delta}=\sqrt{32592}=\sqrt{16*2037}=\sqrt{16}*\sqrt{2037}=4\sqrt{2037}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-180)-4\sqrt{2037}}{2*16}=\frac{180-4\sqrt{2037}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-180)+4\sqrt{2037}}{2*16}=\frac{180+4\sqrt{2037}}{32}
$