-16t^2+40t+6=20

Solution for -16t^2+40t+6=20 equation:

-16t^2+40t+6=20

We move all terms to the left:

-16t^2+40t+6-(20)=0

We add all the numbers together, and all the variables

-16t^2+40t-14=0

a = -16; b = 40; c = -14;
Δ = b2-4ac
Δ = 402-4·(-16)·(-14)
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{11}}{2*-16}=\frac{-40-8\sqrt{11}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{11}}{2*-16}=\frac{-40+8\sqrt{11}}{-32}
$