0=-16t^2+92t+20

Solution for 0=-16t^2+92t+20 equation:

0=-16t^2+92t+20

We move all terms to the left:

0-(-16t^2+92t+20)=0

We add all the numbers together, and all the variables

-(-16t^2+92t+20)=0

We get rid of parentheses

16t^2-92t-20=0

a = 16; b = -92; c = -20;
Δ = b2-4ac
Δ = -922-4·16·(-20)
Δ = 9744
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:

$\sqrt{\Delta}=\sqrt{9744}=\sqrt{16*609}=\sqrt{16}*\sqrt{609}=4\sqrt{609}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-92)-4\sqrt{609}}{2*16}=\frac{92-4\sqrt{609}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-92)+4\sqrt{609}}{2*16}=\frac{92+4\sqrt{609}}{32}
$