0=-4t^2+8t+32

Solution for 0=-4t^2+8t+32 equation:

0=-4t^2+8t+32

We move all terms to the left:

0-(-4t^2+8t+32)=0

We add all the numbers together, and all the variables

-(-4t^2+8t+32)=0

We get rid of parentheses

4t^2-8t-32=0

a = 4; b = -8; c = -32;
Δ = b2-4ac
Δ = -82-4·4·(-32)
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{-(-8)-24}{2*4}=\frac{-16}{8}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+24}{2*4}=\frac{32}{8}
=4
$