0=-16(t+1)(t-4)

Solution for 0=-16(t+1)(t-4) equation:

0=-16(t+1)(t-4)

We move all terms to the left:

0-(-16(t+1)(t-4))=0

We add all the numbers together, and all the variables

-(-16(t+1)(t-4))=0

We multiply parentheses ..

-(-16(+t^2-4t+t-4))=0


We calculate terms in parentheses: -(-16(+t^2-4t+t-4)), so:

-16(+t^2-4t+t-4)

We multiply parentheses

-16t^2+64t-16t+64

We add all the numbers together, and all the variables

-16t^2+48t+64

Back to the equation:

-(-16t^2+48t+64)


We get rid of parentheses

16t^2-48t-64=0

a = 16; b = -48; c = -64;
Δ = b2-4ac
Δ = -482-4·16·(-64)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-80}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+80}{2*16}=\frac{128}{32}
=4
$