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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+t^2+8t-4t-32))+t=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 add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1t^2-3t+32=0

a = -1; b = -3; c = +32;
Δ = b2-4ac
Δ = -32-4·(-1)·32
Δ = 137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{137}}{2*-1}=\frac{3-\sqrt{137}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{137}}{2*-1}=\frac{3+\sqrt{137}}{-2}
$