16t^2+32t+4=0

Solution for 16t^2+32t+4=0 equation:

16t^2+32t+4=0

a = 16; b = 32; c = +4;
Δ = b2-4ac
Δ = 322-4·16·4
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{3}}{2*16}=\frac{-32-16\sqrt{3}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{3}}{2*16}=\frac{-32+16\sqrt{3}}{32}
$