0.8t^2=48t

Solution for 0.8t^2=48t equation:

0.8t^2=48t

We move all terms to the left:

0.8t^2-(48t)=0

a = 0.8; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·0.8·0
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*0.8}=\frac{0}{1.6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*0.8}=\frac{96}{1.6}
=60
$