0=-480t^2+1470t-980

Solution for 0=-480t^2+1470t-980 equation:

0=-480t^2+1470t-980

We move all terms to the left:

0-(-480t^2+1470t-980)=0

We add all the numbers together, and all the variables

-(-480t^2+1470t-980)=0

We get rid of parentheses

480t^2-1470t+980=0

a = 480; b = -1470; c = +980;
Δ = b2-4ac
Δ = -14702-4·480·980
Δ = 279300
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:

$\sqrt{\Delta}=\sqrt{279300}=\sqrt{4900*57}=\sqrt{4900}*\sqrt{57}=70\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1470)-70\sqrt{57}}{2*480}=\frac{1470-70\sqrt{57}}{960}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1470)+70\sqrt{57}}{2*480}=\frac{1470+70\sqrt{57}}{960}
$