0=-16t^2-22t+112

Solution for 0=-16t^2-22t+112 equation:

0=-16t^2-22t+112

We move all terms to the left:

0-(-16t^2-22t+112)=0

We add all the numbers together, and all the variables

-(-16t^2-22t+112)=0

We get rid of parentheses

16t^2+22t-112=0

a = 16; b = 22; c = -112;
Δ = b2-4ac
Δ = 222-4·16·(-112)
Δ = 7652
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:

$\sqrt{\Delta}=\sqrt{7652}=\sqrt{4*1913}=\sqrt{4}*\sqrt{1913}=2\sqrt{1913}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{1913}}{2*16}=\frac{-22-2\sqrt{1913}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{1913}}{2*16}=\frac{-22+2\sqrt{1913}}{32}
$