-16t^2+63t-4=0

Solution for -16t^2+63t-4=0 equation:

-16t^2+63t-4=0

a = -16; b = 63; c = -4;
Δ = b2-4ac
Δ = 632-4·(-16)·(-4)
Δ = 3713
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-\sqrt{3713}}{2*-16}=\frac{-63-\sqrt{3713}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+\sqrt{3713}}{2*-16}=\frac{-63+\sqrt{3713}}{-32}
$