(3+x)*(20+x)=168

Solution for (3+x)*(20+x)=168 equation:

(3+x)(20+x)=168

We move all terms to the left:

(3+x)(20+x)-(168)=0

We add all the numbers together, and all the variables

(x+3)(x+20)-168=0

We multiply parentheses ..

(+x^2+20x+3x+60)-168=0

We get rid of parentheses

x^2+20x+3x+60-168=0

We add all the numbers together, and all the variables

x^2+23x-108=0

a = 1; b = 23; c = -108;
Δ = b2-4ac
Δ = 232-4·1·(-108)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-31}{2*1}=\frac{-54}{2}
=-27
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+31}{2*1}=\frac{8}{2}
=4
$