t^2-13.32t-0.8=0

Solution for t^2-13.32t-0.8=0 equation:

t^2-13.32t-0.8=0

a = 1; b = -13.32; c = -0.8;
Δ = b2-4ac
Δ = -13.322-4·1·(-0.8)
Δ = 180.6224
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{-(-13.32)-\sqrt{180.6224}}{2*1}=\frac{13.32-\sqrt{180.6224}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13.32)+\sqrt{180.6224}}{2*1}=\frac{13.32+\sqrt{180.6224}}{2}
$