16t^2+40t+15=0

Solution for 16t^2+40t+15=0 equation:

16t^2+40t+15=0

a = 16; b = 40; c = +15;
Δ = b2-4ac
Δ = 402-4·16·15
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{10}}{2*16}=\frac{-40-8\sqrt{10}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{10}}{2*16}=\frac{-40+8\sqrt{10}}{32}
$