32=-4t2+24t

Solution for 32=-4t2+24t equation:

32=-4t^2+24t

We move all terms to the left:

32-(-4t^2+24t)=0

We get rid of parentheses

4t^2-24t+32=0

a = 4; b = -24; c = +32;
Δ = b2-4ac
Δ = -242-4·4·32
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{-(-24)-8}{2*4}=\frac{16}{8}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8}{2*4}=\frac{32}{8}
=4
$