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

Solution for (t+5)(t+3)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

t^2+8t+15=0

a = 1; b = 8; c = +15;
Δ = b2-4ac
Δ = 82-4·1·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*1}=\frac{-10}{2}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*1}=\frac{-6}{2}
=-3
$