0=(-16t^2+32t+64)

Solution for 0=(-16t^2+32t+64) equation:

0=(-16t^2+32t+64)

We move all terms to the left:

0-((-16t^2+32t+64))=0

We add all the numbers together, and all the variables

-((-16t^2+32t+64))=0


We calculate terms in parentheses: -((-16t^2+32t+64)), so:

(-16t^2+32t+64)

We get rid of parentheses

-16t^2+32t+64

Back to the equation:

-(-16t^2+32t+64)


We get rid of parentheses

16t^2-32t-64=0

a = 16; b = -32; c = -64;
Δ = b2-4ac
Δ = -322-4·16·(-64)
Δ = 5120
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:

$\sqrt{\Delta}=\sqrt{5120}=\sqrt{1024*5}=\sqrt{1024}*\sqrt{5}=32\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-32\sqrt{5}}{2*16}=\frac{32-32\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+32\sqrt{5}}{2*16}=\frac{32+32\sqrt{5}}{32}
$