16t^2-12t-182=0

Solution for 16t^2-12t-182=0 equation:

16t^2-12t-182=0

a = 16; b = -12; c = -182;
Δ = b2-4ac
Δ = -122-4·16·(-182)
Δ = 11792
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:

$\sqrt{\Delta}=\sqrt{11792}=\sqrt{16*737}=\sqrt{16}*\sqrt{737}=4\sqrt{737}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{737}}{2*16}=\frac{12-4\sqrt{737}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{737}}{2*16}=\frac{12+4\sqrt{737}}{32}
$