t^2+0.08t-0.138=0

Solution for t^2+0.08t-0.138=0 equation:

t^2+0.08t-0.138=0

a = 1; b = 0.08; c = -0.138;
Δ = b2-4ac
Δ = 0.082-4·1·(-0.138)
Δ = 0.5584
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{-(0.08)-\sqrt{0.5584}}{2*1}=\frac{-0.08-\sqrt{0.5584}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.08)+\sqrt{0.5584}}{2*1}=\frac{-0.08+\sqrt{0.5584}}{2}
$