60t^2+120t-30=0

Solution for 60t^2+120t-30=0 equation:

60t^2+120t-30=0

a = 60; b = 120; c = -30;
Δ = b2-4ac
Δ = 1202-4·60·(-30)
Δ = 21600
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:

$\sqrt{\Delta}=\sqrt{21600}=\sqrt{3600*6}=\sqrt{3600}*\sqrt{6}=60\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-60\sqrt{6}}{2*60}=\frac{-120-60\sqrt{6}}{120}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+60\sqrt{6}}{2*60}=\frac{-120+60\sqrt{6}}{120}
$