3t^2-12t+0.2=0

Solution for 3t^2-12t+0.2=0 equation:

3t^2-12t+0.2=0

a = 3; b = -12; c = +0.2;
Δ = b2-4ac
Δ = -122-4·3·0.2
Δ = 141.6
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{-(-12)-\sqrt{141.6}}{2*3}=\frac{12-\sqrt{141.6}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+\sqrt{141.6}}{2*3}=\frac{12+\sqrt{141.6}}{6}
$