16t^2+96t+4=0

Solution for 16t^2+96t+4=0 equation:

16t^2+96t+4=0

a = 16; b = 96; c = +4;
Δ = b2-4ac
Δ = 962-4·16·4
Δ = 8960
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:

$\sqrt{\Delta}=\sqrt{8960}=\sqrt{256*35}=\sqrt{256}*\sqrt{35}=16\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{35}}{2*16}=\frac{-96-16\sqrt{35}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{35}}{2*16}=\frac{-96+16\sqrt{35}}{32}
$