t^2+20.9t-12=0

Solution for t^2+20.9t-12=0 equation:

t^2+20.9t-12=0

a = 1; b = 20.9; c = -12;
Δ = b2-4ac
Δ = 20.92-4·1·(-12)
Δ = 484.81
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{-(20.9)-\sqrt{484.81}}{2*1}=\frac{-20.9-\sqrt{484.81}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20.9)+\sqrt{484.81}}{2*1}=\frac{-20.9+\sqrt{484.81}}{2}
$