0=0.1t^2+3t-400

Solution for 0=0.1t^2+3t-400 equation:

0=0.1t^2+3t-400

We move all terms to the left:

0-(0.1t^2+3t-400)=0

We add all the numbers together, and all the variables

-(0.1t^2+3t-400)=0

We get rid of parentheses

-0.1t^2-3t+400=0

a = -0.1; b = -3; c = +400;
Δ = b2-4ac
Δ = -32-4·(-0.1)·400
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*-0.1}=\frac{-10}{-0.2}
=+50
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*-0.1}=\frac{16}{-0.2}
=-80
$