20=16t2+24t+5

Solution for 20=16t2+24t+5 equation:

20=16t^2+24t+5

We move all terms to the left:

20-(16t^2+24t+5)=0

We get rid of parentheses

-16t^2-24t-5+20=0

We add all the numbers together, and all the variables

-16t^2-24t+15=0

a = -16; b = -24; c = +15;
Δ = b2-4ac
Δ = -242-4·(-16)·15
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16\sqrt{6}}{2*-16}=\frac{24-16\sqrt{6}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16\sqrt{6}}{2*-16}=\frac{24+16\sqrt{6}}{-32}
$