16t^2+29t-123=0

Solution for 16t^2+29t-123=0 equation:

16t^2+29t-123=0

a = 16; b = 29; c = -123;
Δ = b2-4ac
Δ = 292-4·16·(-123)
Δ = 8713
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{-(29)-\sqrt{8713}}{2*16}=\frac{-29-\sqrt{8713}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{8713}}{2*16}=\frac{-29+\sqrt{8713}}{32}
$