-16t^2+8t+7=0

Solution for -16t^2+8t+7=0 equation:

-16t^2+8t+7=0

a = -16; b = 8; c = +7;
Δ = b2-4ac
Δ = 82-4·(-16)·7
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16\sqrt{2}}{2*-16}=\frac{-8-16\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16\sqrt{2}}{2*-16}=\frac{-8+16\sqrt{2}}{-32}
$