3=-16t^2+166t+0

Solution for 3=-16t^2+166t+0 equation:

3=-16t^2+166t+0

We move all terms to the left:

3-(-16t^2+166t+0)=0

We get rid of parentheses

16t^2-166t-0+3=0

We add all the numbers together, and all the variables

16t^2-166t+3=0

a = 16; b = -166; c = +3;
Δ = b2-4ac
Δ = -1662-4·16·3
Δ = 27364
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:

$\sqrt{\Delta}=\sqrt{27364}=\sqrt{4*6841}=\sqrt{4}*\sqrt{6841}=2\sqrt{6841}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-166)-2\sqrt{6841}}{2*16}=\frac{166-2\sqrt{6841}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-166)+2\sqrt{6841}}{2*16}=\frac{166+2\sqrt{6841}}{32}
$