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}
$