0=(3t^2)-40t+128

Solution for 0=(3t^2)-40t+128 equation:

0=(3t^2)-40t+128

We move all terms to the left:

0-((3t^2)-40t+128)=0

We add all the numbers together, and all the variables

-(3t^2-40t+128)=0

We get rid of parentheses

-3t^2+40t-128=0

a = -3; b = 40; c = -128;
Δ = b2-4ac
Δ = 402-4·(-3)·(-128)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8}{2*-3}=\frac{-48}{-6}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8}{2*-3}=\frac{-32}{-6}
=5+1/3
$