48t-18t^2=0

Solution for 48t-18t^2=0 equation:

48t-18t^2=0

a = -18; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·(-18)·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*-18}=\frac{-96}{-36}
=2+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*-18}=\frac{0}{-36}
=0
$