32t-16t^2+13=0

Solution for 32t-16t^2+13=0 equation:

32t-16t^2+13=0

a = -16; b = 32; c = +13;
Δ = b2-4ac
Δ = 322-4·(-16)·13
Δ = 1856
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:

$\sqrt{\Delta}=\sqrt{1856}=\sqrt{64*29}=\sqrt{64}*\sqrt{29}=8\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{29}}{2*-16}=\frac{-32-8\sqrt{29}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{29}}{2*-16}=\frac{-32+8\sqrt{29}}{-32}
$