0=-16t^2+112t+75

Solution for 0=-16t^2+112t+75 equation:

0=-16t^2+112t+75

We move all terms to the left:

0-(-16t^2+112t+75)=0

We add all the numbers together, and all the variables

-(-16t^2+112t+75)=0

We get rid of parentheses

16t^2-112t-75=0

a = 16; b = -112; c = -75;
Δ = b2-4ac
Δ = -1122-4·16·(-75)
Δ = 17344
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:

$\sqrt{\Delta}=\sqrt{17344}=\sqrt{64*271}=\sqrt{64}*\sqrt{271}=8\sqrt{271}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-8\sqrt{271}}{2*16}=\frac{112-8\sqrt{271}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+8\sqrt{271}}{2*16}=\frac{112+8\sqrt{271}}{32}
$