-5t^2+21t+100=0

Solution for -5t^2+21t+100=0 equation:

-5t^2+21t+100=0

a = -5; b = 21; c = +100;
Δ = b2-4ac
Δ = 212-4·(-5)·100
Δ = 2441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{2441}}{2*-5}=\frac{-21-\sqrt{2441}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{2441}}{2*-5}=\frac{-21+\sqrt{2441}}{-10}
$