-5t^2+100=0

Solution for -5t^2+100=0 equation:

-5t^2+100=0

a = -5; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-5)·100
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{5}}{2*-5}=\frac{0-20\sqrt{5}}{-10}
=-\frac{20\sqrt{5}}{-10}
=-\frac{2\sqrt{5}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{5}}{2*-5}=\frac{0+20\sqrt{5}}{-10}
=\frac{20\sqrt{5}}{-10}
=\frac{2\sqrt{5}}{-1}
$