50=-24(t)^2+48(t)+32

Solution for 50=-24(t)^2+48(t)+32 equation:

50=-24(t)^2+48(t)+32

We move all terms to the left:

50-(-24(t)^2+48(t)+32)=0

We get rid of parentheses

24t^2-48t-32+50=0

We add all the numbers together, and all the variables

24t^2-48t+18=0

a = 24; b = -48; c = +18;
Δ = b2-4ac
Δ = -482-4·24·18
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-24}{2*24}=\frac{24}{48}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+24}{2*24}=\frac{72}{48}
=1+1/2
$