16t^2+32t-16=0

Solution for 16t^2+32t-16=0 equation:

16t^2+32t-16=0

a = 16; b = 32; c = -16;
Δ = b2-4ac
Δ = 322-4·16·(-16)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32\sqrt{2}}{2*16}=\frac{-32-32\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32\sqrt{2}}{2*16}=\frac{-32+32\sqrt{2}}{32}
$