16t^2+64t-32=0

Solution for 16t^2+64t-32=0 equation:

16t^2+64t-32=0

a = 16; b = 64; c = -32;
Δ = b2-4ac
Δ = 642-4·16·(-32)
Δ = 6144
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:

$\sqrt{\Delta}=\sqrt{6144}=\sqrt{1024*6}=\sqrt{1024}*\sqrt{6}=32\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-32\sqrt{6}}{2*16}=\frac{-64-32\sqrt{6}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+32\sqrt{6}}{2*16}=\frac{-64+32\sqrt{6}}{32}
$