0=16t(t-3)

Solution for 0=16t(t-3) equation:

0=16t(t-3)

We move all terms to the left:

0-(16t(t-3))=0

We add all the numbers together, and all the variables

-(16t(t-3))=0


We calculate terms in parentheses: -(16t(t-3)), so:

16t(t-3)

We multiply parentheses

16t^2-48t

Back to the equation:

-(16t^2-48t)


We get rid of parentheses

-16t^2+48t=0

a = -16; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·(-16)·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*-16}=\frac{-96}{-32}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*-16}=\frac{0}{-32}
=0
$