13=(t-4)(t-8)

Solution for 13=(t-4)(t-8) equation:

13=(t-4)(t-8)

We move all terms to the left:

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

We multiply parentheses ..

-((+t^2-8t-4t+32))+13=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^2-12t+32

Back to the equation:

-(t^2-12t+32)


We get rid of parentheses

-t^2+12t-32+13=0

We add all the numbers together, and all the variables

-1t^2+12t-19=0

a = -1; b = 12; c = -19;
Δ = b2-4ac
Δ = 122-4·(-1)·(-19)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{17}}{2*-1}=\frac{-12-2\sqrt{17}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{17}}{2*-1}=\frac{-12+2\sqrt{17}}{-2}
$