0=(-16t^2+96t+25)

Solution for 0=(-16t^2+96t+25) equation:

0=(-16t^2+96t+25)

We move all terms to the left:

0-((-16t^2+96t+25))=0

We add all the numbers together, and all the variables

-((-16t^2+96t+25))=0


We calculate terms in parentheses: -((-16t^2+96t+25)), so:

(-16t^2+96t+25)

We get rid of parentheses

-16t^2+96t+25

Back to the equation:

-(-16t^2+96t+25)


We get rid of parentheses

16t^2-96t-25=0

a = 16; b = -96; c = -25;
Δ = b2-4ac
Δ = -962-4·16·(-25)
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-104}{2*16}=\frac{-8}{32}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+104}{2*16}=\frac{200}{32}
=6+1/4
$