16t^2+56t+5=0

Solution for 16t^2+56t+5=0 equation:

16t^2+56t+5=0

a = 16; b = 56; c = +5;
Δ = b2-4ac
Δ = 562-4·16·5
Δ = 2816
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:

$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-16\sqrt{11}}{2*16}=\frac{-56-16\sqrt{11}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+16\sqrt{11}}{2*16}=\frac{-56+16\sqrt{11}}{32}
$