(x+2)(x+4)=20

Solution for (x+2)(x+4)=20 equation:

(x+2)(x+4)=20

We move all terms to the left:

(x+2)(x+4)-(20)=0

We multiply parentheses ..

(+x^2+4x+2x+8)-20=0

We get rid of parentheses

x^2+4x+2x+8-20=0

We add all the numbers together, and all the variables

x^2+6x-12=0

a = 1; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·1·(-12)
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{21}}{2*1}=\frac{-6-2\sqrt{21}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{21}}{2*1}=\frac{-6+2\sqrt{21}}{2}
$