16t^2+24t-6=0

Solution for 16t^2+24t-6=0 equation:

16t^2+24t-6=0

a = 16; b = 24; c = -6;
Δ = b2-4ac
Δ = 242-4·16·(-6)
Δ = 960
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:

$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{15}}{2*16}=\frac{-24-8\sqrt{15}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{15}}{2*16}=\frac{-24+8\sqrt{15}}{32}
$