(t+4)(t+2)=3

Solution for (t+4)(t+2)=3 equation:

(t+4)(t+2)=3

We move all terms to the left:

(t+4)(t+2)-(3)=0

We multiply parentheses ..

(+t^2+2t+4t+8)-3=0

We get rid of parentheses

t^2+2t+4t+8-3=0

We add all the numbers together, and all the variables

t^2+6t+5=0

a = 1; b = 6; c = +5;
Δ = b2-4ac
Δ = 62-4·1·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*1}=\frac{-10}{2}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*1}=\frac{-2}{2}
=-1
$