16t^2+24t+5=0

Solution for 16t^2+24t+5=0 equation:

16t^2+24t+5=0

a = 16; b = 24; c = +5;
Δ = b2-4ac
Δ = 242-4·16·5
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16}{2*16}=\frac{-40}{32}
=-1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16}{2*16}=\frac{-8}{32}
=-1/4
$