16t^2+9t-171=0

Solution for 16t^2+9t-171=0 equation:

16t^2+9t-171=0

a = 16; b = 9; c = -171;
Δ = b2-4ac
Δ = 92-4·16·(-171)
Δ = 11025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11025}=105$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-105}{2*16}=\frac{-114}{32}
=-3+9/16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+105}{2*16}=\frac{96}{32}
=3
$