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

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

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