0=-16t^2+92t+140

Solution for 0=-16t^2+92t+140 equation:

0=-16t^2+92t+140

We move all terms to the left:

0-(-16t^2+92t+140)=0

We add all the numbers together, and all the variables

-(-16t^2+92t+140)=0

We get rid of parentheses

16t^2-92t-140=0

a = 16; b = -92; c = -140;
Δ = b2-4ac
Δ = -922-4·16·(-140)
Δ = 17424
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{17424}=132$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-92)-132}{2*16}=\frac{-40}{32}
=-1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-92)+132}{2*16}=\frac{224}{32}
=7
$