0=0.1t^2+3t

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

0=0.1t^2+3t

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-0.1t^2-3t=0

a = -0.1; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-0.1)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-0.1}=\frac{0}{-0.2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-0.1}=\frac{6}{-0.2}
=-30
$