0=-16t^2-96t+7

Solution for 0=-16t^2-96t+7 equation:

0=-16t^2-96t+7

We move all terms to the left:

0-(-16t^2-96t+7)=0

We add all the numbers together, and all the variables

-(-16t^2-96t+7)=0

We get rid of parentheses

16t^2+96t-7=0

a = 16; b = 96; c = -7;
Δ = b2-4ac
Δ = 962-4·16·(-7)
Δ = 9664
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:

$\sqrt{\Delta}=\sqrt{9664}=\sqrt{64*151}=\sqrt{64}*\sqrt{151}=8\sqrt{151}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-8\sqrt{151}}{2*16}=\frac{-96-8\sqrt{151}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+8\sqrt{151}}{2*16}=\frac{-96+8\sqrt{151}}{32}
$