(10+x)(12+x)=240

Solution for (10+x)(12+x)=240 equation:

(10+x)(12+x)=240

We move all terms to the left:

(10+x)(12+x)-(240)=0

We add all the numbers together, and all the variables

(x+10)(x+12)-240=0

We multiply parentheses ..

(+x^2+12x+10x+120)-240=0

We get rid of parentheses

x^2+12x+10x+120-240=0

We add all the numbers together, and all the variables

x^2+22x-120=0

a = 1; b = 22; c = -120;
Δ = b2-4ac
Δ = 222-4·1·(-120)
Δ = 964
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:

$\sqrt{\Delta}=\sqrt{964}=\sqrt{4*241}=\sqrt{4}*\sqrt{241}=2\sqrt{241}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{241}}{2*1}=\frac{-22-2\sqrt{241}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{241}}{2*1}=\frac{-22+2\sqrt{241}}{2}
$