-480+30t^2=-400

Solution for -480+30t^2=-400 equation:

-480+30t^2=-400

We move all terms to the left:

-480+30t^2-(-400)=0

We add all the numbers together, and all the variables

30t^2-80=0

a = 30; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·30·(-80)
Δ = 9600
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:

$\sqrt{\Delta}=\sqrt{9600}=\sqrt{1600*6}=\sqrt{1600}*\sqrt{6}=40\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{6}}{2*30}=\frac{0-40\sqrt{6}}{60}
=-\frac{40\sqrt{6}}{60}
=-\frac{2\sqrt{6}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{6}}{2*30}=\frac{0+40\sqrt{6}}{60}
=\frac{40\sqrt{6}}{60}
=\frac{2\sqrt{6}}{3}
$