0=16t^2+50t+36

Solution for 0=16t^2+50t+36 equation:

0=16t^2+50t+36

We move all terms to the left:

0-(16t^2+50t+36)=0

We add all the numbers together, and all the variables

-(16t^2+50t+36)=0

We get rid of parentheses

-16t^2-50t-36=0

a = -16; b = -50; c = -36;
Δ = b2-4ac
Δ = -502-4·(-16)·(-36)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-14}{2*-16}=\frac{36}{-32}
=-1+1/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+14}{2*-16}=\frac{64}{-32}
=-2
$