-5t^2+30t+200=0

Solution for -5t^2+30t+200=0 equation:

-5t^2+30t+200=0

a = -5; b = 30; c = +200;
Δ = b2-4ac
Δ = 302-4·(-5)·200
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-70}{2*-5}=\frac{-100}{-10}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+70}{2*-5}=\frac{40}{-10}
=-4
$