t^2-0.4t-0.6=0

Solution for t^2-0.4t-0.6=0 equation:

t^2-0.4t-0.6=0

a = 1; b = -0.4; c = -0.6;
Δ = b2-4ac
Δ = -0.42-4·1·(-0.6)
Δ = 2.56
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{-(-0.4)-\sqrt{2.56}}{2*1}=\frac{0.4-\sqrt{2.56}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.4)+\sqrt{2.56}}{2*1}=\frac{0.4+\sqrt{2.56}}{2}
$