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

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

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

We move all terms to the left:

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

determiningTheFunctionDomain
-16t^2+40t-20t+20=0

We add all the numbers together, and all the variables

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

a = -16; b = 20; c = +20;
Δ = b2-4ac
Δ = 202-4·(-16)·20
Δ = 1680
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:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{105}}{2*-16}=\frac{-20-4\sqrt{105}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{105}}{2*-16}=\frac{-20+4\sqrt{105}}{-32}
$