-30t^2+410t-330=0

Solution for -30t^2+410t-330=0 equation:

-30t^2+410t-330=0

a = -30; b = 410; c = -330;
Δ = b2-4ac
Δ = 4102-4·(-30)·(-330)
Δ = 128500
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:

$\sqrt{\Delta}=\sqrt{128500}=\sqrt{100*1285}=\sqrt{100}*\sqrt{1285}=10\sqrt{1285}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(410)-10\sqrt{1285}}{2*-30}=\frac{-410-10\sqrt{1285}}{-60}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(410)+10\sqrt{1285}}{2*-30}=\frac{-410+10\sqrt{1285}}{-60}
$