-15(t)^2+300(t)-200=0

Solution for -15(t)^2+300(t)-200=0 equation:

-15(t)^2+300(t)-200=0

a = -15; b = 300; c = -200;
Δ = b2-4ac
Δ = 3002-4·(-15)·(-200)
Δ = 78000
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:

$\sqrt{\Delta}=\sqrt{78000}=\sqrt{400*195}=\sqrt{400}*\sqrt{195}=20\sqrt{195}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(300)-20\sqrt{195}}{2*-15}=\frac{-300-20\sqrt{195}}{-30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(300)+20\sqrt{195}}{2*-15}=\frac{-300+20\sqrt{195}}{-30}
$