-0.2t^2+2.4t-6=0

Solution for -0.2t^2+2.4t-6=0 equation:

-0.2t^2+2.4t-6=0

a = -0.2; b = 2.4; c = -6;
Δ = b2-4ac
Δ = 2.42-4·(-0.2)·(-6)
Δ = 0.96
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{-(2.4)-\sqrt{0.96}}{2*-0.2}=\frac{-2.4-\sqrt{0.96}}{-0.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.4)+\sqrt{0.96}}{2*-0.2}=\frac{-2.4+\sqrt{0.96}}{-0.4}
$