16t2+6t-130=0

Solution for 16t2+6t-130=0 equation:

16t^2+6t-130=0

a = 16; b = 6; c = -130;
Δ = b2-4ac
Δ = 62-4·16·(-130)
Δ = 8356
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:

$\sqrt{\Delta}=\sqrt{8356}=\sqrt{4*2089}=\sqrt{4}*\sqrt{2089}=2\sqrt{2089}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{2089}}{2*16}=\frac{-6-2\sqrt{2089}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{2089}}{2*16}=\frac{-6+2\sqrt{2089}}{32}
$