t^2-4t-0.4=0

Solution for t^2-4t-0.4=0 equation:

t^2-4t-0.4=0

a = 1; b = -4; c = -0.4;
Δ = b2-4ac
Δ = -42-4·1·(-0.4)
Δ = 17.6
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{-(-4)-\sqrt{17.6}}{2*1}=\frac{4-\sqrt{17.6}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{17.6}}{2*1}=\frac{4+\sqrt{17.6}}{2}
$