100+20t-3t^2=0

Solution for 100+20t-3t^2=0 equation:

100+20t-3t^2=0

a = -3; b = 20; c = +100;
Δ = b2-4ac
Δ = 202-4·(-3)·100
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-40}{2*-3}=\frac{-60}{-6}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+40}{2*-3}=\frac{20}{-6}
=-3+1/3
$