0=(t-4)(t+8)

Solution for 0=(t-4)(t+8) equation:

0=(t-4)(t+8)

We move all terms to the left:

0-((t-4)(t+8))=0

We add all the numbers together, and all the variables

-((t-4)(t+8))=0

We multiply parentheses ..

-((+t^2+8t-4t-32))=0


We calculate terms in parentheses: -((+t^2+8t-4t-32)), so:

(+t^2+8t-4t-32)

We get rid of parentheses

t^2+8t-4t-32

We add all the numbers together, and all the variables

t^2+4t-32

Back to the equation:

-(t^2+4t-32)


We get rid of parentheses

-t^2-4t+32=0

We add all the numbers together, and all the variables

-1t^2-4t+32=0

a = -1; b = -4; c = +32;
Δ = b2-4ac
Δ = -42-4·(-1)·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*-1}=\frac{-8}{-2}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*-1}=\frac{16}{-2}
=-8
$