(t+5)(t+3)=1

Solution for (t+5)(t+3)=1 equation:

(t+5)(t+3)=1

We move all terms to the left:

(t+5)(t+3)-(1)=0

We multiply parentheses ..

(+t^2+3t+5t+15)-1=0

We get rid of parentheses

t^2+3t+5t+15-1=0

We add all the numbers together, and all the variables

t^2+8t+14=0

a = 1; b = 8; c = +14;
Δ = b2-4ac
Δ = 82-4·1·14
Δ = 8
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:

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