5.4t-0.36t^2=0

Solution for 5.4t-0.36t^2=0 equation:

5.4t-0.36t^2=0

a = -0.36; b = 5.4; c = 0;
Δ = b2-4ac
Δ = 5.42-4·(-0.36)·0
Δ = 29.16
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{-(5.4)-\sqrt{29.16}}{2*-0.36}=\frac{-5.4-\sqrt{29.16}}{-0.72}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.4)+\sqrt{29.16}}{2*-0.36}=\frac{-5.4+\sqrt{29.16}}{-0.72}
$