-4t^2+8t=-32

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

-4t^2+8t=-32

We move all terms to the left:

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

We add all the numbers together, and all the variables

-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{-32}{-8}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+24}{2*-4}=\frac{16}{-8}
=-2
$