-16t^2+56-20=0

Solution for -16t^2+56-20=0 equation:

-16t^2+56-20=0

We add all the numbers together, and all the variables

-16t^2+36=0

a = -16; b = 0; c = +36;
Δ = b2-4ac
Δ = 02-4·(-16)·36
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*-16}=\frac{-48}{-32}
=1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*-16}=\frac{48}{-32}
=-1+1/2
$