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

Solution for 0(t)=-16t^2+520 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2+t-520=0

a = 16; b = 1; c = -520;
Δ = b2-4ac
Δ = 12-4·16·(-520)
Δ = 33281
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{-(1)-\sqrt{33281}}{2*16}=\frac{-1-\sqrt{33281}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{33281}}{2*16}=\frac{-1+\sqrt{33281}}{32}
$