-16t^2+16t-3.5=0

Solution for -16t^2+16t-3.5=0 equation:

-16t^2+16t-3.5=0

a = -16; b = 16; c = -3.5;
Δ = b2-4ac
Δ = 162-4·(-16)·(-3.5)
Δ = 32
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:

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