(t-4)(t+2)=1

Solution for (t-4)(t+2)=1 equation:

(t-4)(t+2)=1

We move all terms to the left:

(t-4)(t+2)-(1)=0

We multiply parentheses ..

(+t^2+2t-4t-8)-1=0

We get rid of parentheses

t^2+2t-4t-8-1=0

We add all the numbers together, and all the variables

t^2-2t-9=0

a = 1; b = -2; c = -9;
Δ = b2-4ac
Δ = -22-4·1·(-9)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{10}}{2*1}=\frac{2-2\sqrt{10}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{10}}{2*1}=\frac{2+2\sqrt{10}}{2}
$