15=-16t^2+32t+6

Solution for 15=-16t^2+32t+6 equation:

15=-16t^2+32t+6

We move all terms to the left:

15-(-16t^2+32t+6)=0

We get rid of parentheses

16t^2-32t-6+15=0

We add all the numbers together, and all the variables

16t^2-32t+9=0

a = 16; b = -32; c = +9;
Δ = b2-4ac
Δ = -322-4·16·9
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{7}}{2*16}=\frac{32-8\sqrt{7}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{7}}{2*16}=\frac{32+8\sqrt{7}}{32}
$