16t^2+160t-110=0

Solution for 16t^2+160t-110=0 equation:

16t^2+160t-110=0

a = 16; b = 160; c = -110;
Δ = b2-4ac
Δ = 1602-4·16·(-110)
Δ = 32640
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:

$\sqrt{\Delta}=\sqrt{32640}=\sqrt{64*510}=\sqrt{64}*\sqrt{510}=8\sqrt{510}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-8\sqrt{510}}{2*16}=\frac{-160-8\sqrt{510}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+8\sqrt{510}}{2*16}=\frac{-160+8\sqrt{510}}{32}
$