0.36t2+4.8t=0

Solution for 0.36t2+4.8t=0 equation:

0.36t^2+4.8t=0

a = 0.36; b = 4.8; c = 0;
Δ = b2-4ac
Δ = 4.82-4·0.36·0
Δ = 23.04
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{-(4.8)-\sqrt{23.04}}{2*0.36}=\frac{-4.8-\sqrt{23.04}}{0.72}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.8)+\sqrt{23.04}}{2*0.36}=\frac{-4.8+\sqrt{23.04}}{0.72}
$