t^2-0,5*t-0,6=0

Solution for t^2-0,5*t-0,6=0 equation:

t^2-0.5t-0.6=0

a = 1; b = -0.5; c = -0.6;
Δ = b2-4ac
Δ = -0.52-4·1·(-0.6)
Δ = 2.65
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{-(-0.5)-\sqrt{2.65}}{2*1}=\frac{0.5-\sqrt{2.65}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{2.65}}{2*1}=\frac{0.5+\sqrt{2.65}}{2}
$