(3+x)(x-5)=20

Solution for (3+x)(x-5)=20 equation:

(3+x)(x-5)=20

We move all terms to the left:

(3+x)(x-5)-(20)=0

We add all the numbers together, and all the variables

(x+3)(x-5)-20=0

We multiply parentheses ..

(+x^2-5x+3x-15)-20=0

We get rid of parentheses

x^2-5x+3x-15-20=0

We add all the numbers together, and all the variables

x^2-2x-35=0

a = 1; b = -2; c = -35;
Δ = b2-4ac
Δ = -22-4·1·(-35)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-12}{2*1}=\frac{-10}{2}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+12}{2*1}=\frac{14}{2}
=7
$