32=32-2t-5t^2

Solution for 32=32-2t-5t^2 equation:

32=32-2t-5t^2

We move all terms to the left:

32-(32-2t-5t^2)=0

We get rid of parentheses

5t^2+2t-32+32=0

We add all the numbers together, and all the variables

5t^2+2t=0

a = 5; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·5·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*5}=\frac{-4}{10}
=-2/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*5}=\frac{0}{10}
=0
$