4t-0,48t^2=0

Solution for 4t-0,48t^2=0 equation:

4t-0.48t^2=0

a = -0.48; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-0.48)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-0.48}=\frac{-8}{-0.96}
=8+0.32/0.96
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-0.48}=\frac{0}{-0.96}
=0
$