2.7t^2+50t-5.5=0

Solution for 2.7t^2+50t-5.5=0 equation:

2.7t^2+50t-5.5=0

a = 2.7; b = 50; c = -5.5;
Δ = b2-4ac
Δ = 502-4·2.7·(-5.5)
Δ = 2559.4
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{-(50)-\sqrt{2559.4}}{2*2.7}=\frac{-50-\sqrt{2559.4}}{5.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+\sqrt{2559.4}}{2*2.7}=\frac{-50+\sqrt{2559.4}}{5.4}
$