0.1t^2+5.8334t-200=0

Solution for 0.1t^2+5.8334t-200=0 equation:

0.1t^2+5.8334t-200=0

a = 0.1; b = 5.8334; c = -200;
Δ = b2-4ac
Δ = 5.83342-4·0.1·(-200)
Δ = 114.02855556
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{-(5.8334)-\sqrt{114.02855556}}{2*0.1}=\frac{-5.8334-\sqrt{114.02855556}}{0.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.8334)+\sqrt{114.02855556}}{2*0.1}=\frac{-5.8334+\sqrt{114.02855556}}{0.2}
$