130=16t^2+112t+6

Solution for 130=16t^2+112t+6 equation:

130=16t^2+112t+6

We move all terms to the left:

130-(16t^2+112t+6)=0

We get rid of parentheses

-16t^2-112t-6+130=0

We add all the numbers together, and all the variables

-16t^2-112t+124=0

a = -16; b = -112; c = +124;
Δ = b2-4ac
Δ = -1122-4·(-16)·124
Δ = 20480
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:

$\sqrt{\Delta}=\sqrt{20480}=\sqrt{4096*5}=\sqrt{4096}*\sqrt{5}=64\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-64\sqrt{5}}{2*-16}=\frac{112-64\sqrt{5}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+64\sqrt{5}}{2*-16}=\frac{112+64\sqrt{5}}{-32}
$