32=2t+5t^2

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

32=2t+5t^2

We move all terms to the left:

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

We get rid of parentheses

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

a = -5; b = -2; c = +32;
Δ = b2-4ac
Δ = -22-4·(-5)·32
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{161}}{2*-5}=\frac{2-2\sqrt{161}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{161}}{2*-5}=\frac{2+2\sqrt{161}}{-10}
$