16t^2+28t+8=0

Solution for 16t^2+28t+8=0 equation:

16t^2+28t+8=0

a = 16; b = 28; c = +8;
Δ = b2-4ac
Δ = 282-4·16·8
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{17}}{2*16}=\frac{-28-4\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{17}}{2*16}=\frac{-28+4\sqrt{17}}{32}
$