540=(40+2x)(60+x)+x

Solution for 540=(40+2x)(60+x)+x equation:

540=(40+2x)(60+x)+x

We move all terms to the left:

540-((40+2x)(60+x)+x)=0

We add all the numbers together, and all the variables

-((2x+40)(x+60)+x)+540=0

We multiply parentheses ..

-((+2x^2+120x+40x+2400)+x)+540=0


We calculate terms in parentheses: -((+2x^2+120x+40x+2400)+x), so:

(+2x^2+120x+40x+2400)+x

We get rid of parentheses

2x^2+120x+40x+x+2400

We add all the numbers together, and all the variables

2x^2+161x+2400

Back to the equation:

-(2x^2+161x+2400)


We get rid of parentheses

-2x^2-161x-2400+540=0

We add all the numbers together, and all the variables

-2x^2-161x-1860=0

a = -2; b = -161; c = -1860;
Δ = b2-4ac
Δ = -1612-4·(-2)·(-1860)
Δ = 11041
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{-(-161)-\sqrt{11041}}{2*-2}=\frac{161-\sqrt{11041}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-161)+\sqrt{11041}}{2*-2}=\frac{161+\sqrt{11041}}{-4}
$