-130=-16t^2+4t

Solution for -130=-16t^2+4t equation:

-130=-16t^2+4t

We move all terms to the left:

-130-(-16t^2+4t)=0

We get rid of parentheses

16t^2-4t-130=0

a = 16; b = -4; c = -130;
Δ = b2-4ac
Δ = -42-4·16·(-130)
Δ = 8336
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:

$\sqrt{\Delta}=\sqrt{8336}=\sqrt{16*521}=\sqrt{16}*\sqrt{521}=4\sqrt{521}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{521}}{2*16}=\frac{4-4\sqrt{521}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{521}}{2*16}=\frac{4+4\sqrt{521}}{32}
$