162t=-16t^2+162(0)

Solution for 162t=-16t^2+162(0) equation:

162t=-16t^2+162(0)

We move all terms to the left:

162t-(-16t^2+162(0))=0

We get rid of parentheses

16t^2+162t-1620=0

a = 16; b = 162; c = -1620;
Δ = b2-4ac
Δ = 1622-4·16·(-1620)
Δ = 129924
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:

$\sqrt{\Delta}=\sqrt{129924}=\sqrt{324*401}=\sqrt{324}*\sqrt{401}=18\sqrt{401}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(162)-18\sqrt{401}}{2*16}=\frac{-162-18\sqrt{401}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(162)+18\sqrt{401}}{2*16}=\frac{-162+18\sqrt{401}}{32}
$