(x+3)(x+2)=20

Solution for (x+3)(x+2)=20 equation:

(x+3)(x+2)=20

We move all terms to the left:

(x+3)(x+2)-(20)=0

We multiply parentheses ..

(+x^2+2x+3x+6)-20=0

We get rid of parentheses

x^2+2x+3x+6-20=0

We add all the numbers together, and all the variables

x^2+5x-14=0

a = 1; b = 5; c = -14;
Δ = b2-4ac
Δ = 52-4·1·(-14)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-9}{2*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+9}{2*1}=\frac{4}{2}
=2
$