-.3t^2+18t=0

Solution for -.3t^2+18t=0 equation:

-.3t^2+18t=0

We add all the numbers together, and all the variables

-0.3t^2+18t=0

a = -0.3; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·(-0.3)·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*-0.3}=\frac{-36}{-0.6}
=+60
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*-0.3}=\frac{0}{-0.6}
=0
$