0=-16t2+20t+50

Solution for 0=-16t2+20t+50 equation:

0=-16t^2+20t+50

We move all terms to the left:

0-(-16t^2+20t+50)=0

We add all the numbers together, and all the variables

-(-16t^2+20t+50)=0

We get rid of parentheses

16t^2-20t-50=0

a = 16; b = -20; c = -50;
Δ = b2-4ac
Δ = -202-4·16·(-50)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-60}{2*16}=\frac{-40}{32}
=-1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+60}{2*16}=\frac{80}{32}
=2+1/2
$