-12t^2+480t+30=0

Solution for -12t^2+480t+30=0 equation:

-12t^2+480t+30=0

a = -12; b = 480; c = +30;
Δ = b2-4ac
Δ = 4802-4·(-12)·30
Δ = 231840
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:

$\sqrt{\Delta}=\sqrt{231840}=\sqrt{144*1610}=\sqrt{144}*\sqrt{1610}=12\sqrt{1610}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(480)-12\sqrt{1610}}{2*-12}=\frac{-480-12\sqrt{1610}}{-24}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(480)+12\sqrt{1610}}{2*-12}=\frac{-480+12\sqrt{1610}}{-24}
$