-16t^2+40*t+4=4

Solution for -16t^2+40*t+4=4 equation:

-16t^2+40t+4=4

We move all terms to the left:

-16t^2+40t+4-(4)=0

We add all the numbers together, and all the variables

-16t^2+40t=0

a = -16; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·(-16)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*-16}=\frac{-80}{-32}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*-16}=\frac{0}{-32}
=0
$