16t^2+40t+5=0

Solution for 16t^2+40t+5=0 equation:

16t^2+40t+5=0

a = 16; b = 40; c = +5;
Δ = b2-4ac
Δ = 402-4·16·5
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16\sqrt{5}}{2*16}=\frac{-40-16\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{5}}{2*16}=\frac{-40+16\sqrt{5}}{32}
$