0.04t^2+2t-200=0

Solution for 0.04t^2+2t-200=0 equation:

0.04t^2+2t-200=0

a = 0.04; b = 2; c = -200;
Δ = b2-4ac
Δ = 22-4·0.04·(-200)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*0.04}=\frac{-8}{0.08}
=-100
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*0.04}=\frac{4}{0.08}
=50
$