200=0.1(t^2)+3t

Solution for 200=0.1(t^2)+3t equation:

200=0.1(t^2)+3t

We move all terms to the left:

200-(0.1(t^2)+3t)=0

We get rid of parentheses

-0.1t^2-3t+200=0

a = -0.1; b = -3; c = +200;
Δ = b2-4ac
Δ = -32-4·(-0.1)·200
Δ = 89
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{-(-3)-\sqrt{89}}{2*-0.1}=\frac{3-\sqrt{89}}{-0.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*-0.1}=\frac{3+\sqrt{89}}{-0.2}
$