-16t^2+36t-20=0

Solution for -16t^2+36t-20=0 equation:

-16t^2+36t-20=0

a = -16; b = 36; c = -20;
Δ = b2-4ac
Δ = 362-4·(-16)·(-20)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4}{2*-16}=\frac{-40}{-32}
=1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4}{2*-16}=\frac{-32}{-32}
=1
$