.9t^2-3.28t-5.89=0

Solution for .9t^2-3.28t-5.89=0 equation:

.9t^2-3.28t-5.89=0

a = .9; b = -3.28; c = -5.89;
Δ = b2-4ac
Δ = -3.282-4·.9·(-5.89)
Δ = 31.9624
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{-(-3.28)-\sqrt{31.9624}}{2*.9}=\frac{3.28-\sqrt{31.9624}}{1.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3.28)+\sqrt{31.9624}}{2*.9}=\frac{3.28+\sqrt{31.9624}}{1.8}
$