0=16t^2+144t+20

Solution for 0=16t^2+144t+20 equation:

0=16t^2+144t+20

We move all terms to the left:

0-(16t^2+144t+20)=0

We add all the numbers together, and all the variables

-(16t^2+144t+20)=0

We get rid of parentheses

-16t^2-144t-20=0

a = -16; b = -144; c = -20;
Δ = b2-4ac
Δ = -1442-4·(-16)·(-20)
Δ = 19456
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:

$\sqrt{\Delta}=\sqrt{19456}=\sqrt{1024*19}=\sqrt{1024}*\sqrt{19}=32\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-144)-32\sqrt{19}}{2*-16}=\frac{144-32\sqrt{19}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-144)+32\sqrt{19}}{2*-16}=\frac{144+32\sqrt{19}}{-32}
$