-520=-16t^2+35t

Solution for -520=-16t^2+35t equation:

-520=-16t^2+35t

We move all terms to the left:

-520-(-16t^2+35t)=0

We get rid of parentheses

16t^2-35t-520=0

a = 16; b = -35; c = -520;
Δ = b2-4ac
Δ = -352-4·16·(-520)
Δ = 34505
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{34505}}{2*16}=\frac{35-\sqrt{34505}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{34505}}{2*16}=\frac{35+\sqrt{34505}}{32}
$