16t^2+6t-110=0

Solution for 16t^2+6t-110=0 equation:

16t^2+6t-110=0

a = 16; b = 6; c = -110;
Δ = b2-4ac
Δ = 62-4·16·(-110)
Δ = 7076
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:

$\sqrt{\Delta}=\sqrt{7076}=\sqrt{4*1769}=\sqrt{4}*\sqrt{1769}=2\sqrt{1769}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{1769}}{2*16}=\frac{-6-2\sqrt{1769}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{1769}}{2*16}=\frac{-6+2\sqrt{1769}}{32}
$