(4+x)(5+x)=40

Solution for (4+x)(5+x)=40 equation:

(4+x)(5+x)=40

We move all terms to the left:

(4+x)(5+x)-(40)=0

We add all the numbers together, and all the variables

(x+4)(x+5)-40=0

We multiply parentheses ..

(+x^2+5x+4x+20)-40=0

We get rid of parentheses

x^2+5x+4x+20-40=0

We add all the numbers together, and all the variables

x^2+9x-20=0

a = 1; b = 9; c = -20;
Δ = b2-4ac
Δ = 92-4·1·(-20)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{161}}{2*1}=\frac{-9-\sqrt{161}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{161}}{2*1}=\frac{-9+\sqrt{161}}{2}
$