0=-0.2t^2+16t+12

Solution for 0=-0.2t^2+16t+12 equation:

0=-0.2t^2+16t+12

We move all terms to the left:

0-(-0.2t^2+16t+12)=0

We add all the numbers together, and all the variables

-(-0.2t^2+16t+12)=0

We get rid of parentheses

0.2t^2-16t-12=0

a = 0.2; b = -16; c = -12;
Δ = b2-4ac
Δ = -162-4·0.2·(-12)
Δ = 265.6
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{-(-16)-\sqrt{265.6}}{2*0.2}=\frac{16-\sqrt{265.6}}{0.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+\sqrt{265.6}}{2*0.2}=\frac{16+\sqrt{265.6}}{0.4}
$