-5t^2+100t+10=0

Solution for -5t^2+100t+10=0 equation:

-5t^2+100t+10=0

a = -5; b = 100; c = +10;
Δ = b2-4ac
Δ = 1002-4·(-5)·10
Δ = 10200
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:

$\sqrt{\Delta}=\sqrt{10200}=\sqrt{100*102}=\sqrt{100}*\sqrt{102}=10\sqrt{102}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-10\sqrt{102}}{2*-5}=\frac{-100-10\sqrt{102}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+10\sqrt{102}}{2*-5}=\frac{-100+10\sqrt{102}}{-10}
$