0=-16t^2+112t-2

Solution for 0=-16t^2+112t-2 equation:

0=-16t^2+112t-2

We move all terms to the left:

0-(-16t^2+112t-2)=0

We add all the numbers together, and all the variables

-(-16t^2+112t-2)=0

We get rid of parentheses

16t^2-112t+2=0

a = 16; b = -112; c = +2;
Δ = b2-4ac
Δ = -1122-4·16·2
Δ = 12416
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:

$\sqrt{\Delta}=\sqrt{12416}=\sqrt{64*194}=\sqrt{64}*\sqrt{194}=8\sqrt{194}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-8\sqrt{194}}{2*16}=\frac{112-8\sqrt{194}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+8\sqrt{194}}{2*16}=\frac{112+8\sqrt{194}}{32}
$