-16t2+36t-8=0

Solution for -16t2+36t-8=0 equation:

-16t^2+36t-8=0

a = -16; b = 36; c = -8;
Δ = b2-4ac
Δ = 362-4·(-16)·(-8)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-28}{2*-16}=\frac{-64}{-32}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+28}{2*-16}=\frac{-8}{-32}
=1/4
$