-5t^2+500t-1000=0

Solution for -5t^2+500t-1000=0 equation:

-5t^2+500t-1000=0

a = -5; b = 500; c = -1000;
Δ = b2-4ac
Δ = 5002-4·(-5)·(-1000)
Δ = 230000
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:

$\sqrt{\Delta}=\sqrt{230000}=\sqrt{10000*23}=\sqrt{10000}*\sqrt{23}=100\sqrt{23}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(500)-100\sqrt{23}}{2*-5}=\frac{-500-100\sqrt{23}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(500)+100\sqrt{23}}{2*-5}=\frac{-500+100\sqrt{23}}{-10}
$