256-21t-16t^2=0

Solution for 256-21t-16t^2=0 equation:

256-21t-16t^2=0

a = -16; b = -21; c = +256;
Δ = b2-4ac
Δ = -212-4·(-16)·256
Δ = 16825
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{16825}=\sqrt{25*673}=\sqrt{25}*\sqrt{673}=5\sqrt{673}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-5\sqrt{673}}{2*-16}=\frac{21-5\sqrt{673}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+5\sqrt{673}}{2*-16}=\frac{21+5\sqrt{673}}{-32}
$