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

Solution for 0=-5t^2+32t+2 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

5t^2-32t-2=0

a = 5; b = -32; c = -2;
Δ = b2-4ac
Δ = -322-4·5·(-2)
Δ = 1064
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1064}=\sqrt{4*266}=\sqrt{4}*\sqrt{266}=2\sqrt{266}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-2\sqrt{266}}{2*5}=\frac{32-2\sqrt{266}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+2\sqrt{266}}{2*5}=\frac{32+2\sqrt{266}}{10}
$