0=32t^2+40t-48

Solution for 0=32t^2+40t-48 equation:

0=32t^2+40t-48

We move all terms to the left:

0-(32t^2+40t-48)=0

We add all the numbers together, and all the variables

-(32t^2+40t-48)=0

We get rid of parentheses

-32t^2-40t+48=0

a = -32; b = -40; c = +48;
Δ = b2-4ac
Δ = -402-4·(-32)·48
Δ = 7744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7744}=88$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-88}{2*-32}=\frac{-48}{-64}
=3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+88}{2*-32}=\frac{128}{-64}
=-2
$