16t^2+3t-168=0

Solution for 16t^2+3t-168=0 equation:

16t^2+3t-168=0

a = 16; b = 3; c = -168;
Δ = b2-4ac
Δ = 32-4·16·(-168)
Δ = 10761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{10761}}{2*16}=\frac{-3-\sqrt{10761}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{10761}}{2*16}=\frac{-3+\sqrt{10761}}{32}
$