5+32t-16t^2=0

Solution for 5+32t-16t^2=0 equation:

5+32t-16t^2=0

a = -16; b = 32; c = +5;
Δ = b2-4ac
Δ = 322-4·(-16)·5
Δ = 1344
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:

$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{21}}{2*-16}=\frac{-32-8\sqrt{21}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{21}}{2*-16}=\frac{-32+8\sqrt{21}}{-32}
$