20=-16t^2+44t+12

Solution for 20=-16t^2+44t+12 equation:

20=-16t^2+44t+12

We move all terms to the left:

20-(-16t^2+44t+12)=0

We get rid of parentheses

16t^2-44t-12+20=0

We add all the numbers together, and all the variables

16t^2-44t+8=0

a = 16; b = -44; c = +8;
Δ = b2-4ac
Δ = -442-4·16·8
Δ = 1424
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:

$\sqrt{\Delta}=\sqrt{1424}=\sqrt{16*89}=\sqrt{16}*\sqrt{89}=4\sqrt{89}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-4\sqrt{89}}{2*16}=\frac{44-4\sqrt{89}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+4\sqrt{89}}{2*16}=\frac{44+4\sqrt{89}}{32}
$