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

Solution for 0=(40+2x)(60+2x) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+4x^2+120x+80x+2400))=0


We calculate terms in parentheses: -((+4x^2+120x+80x+2400)), so:

(+4x^2+120x+80x+2400)

We get rid of parentheses

4x^2+120x+80x+2400

We add all the numbers together, and all the variables

4x^2+200x+2400

Back to the equation:

-(4x^2+200x+2400)


We get rid of parentheses

-4x^2-200x-2400=0

a = -4; b = -200; c = -2400;
Δ = b2-4ac
Δ = -2002-4·(-4)·(-2400)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-200)-40}{2*-4}=\frac{160}{-8}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-200)+40}{2*-4}=\frac{240}{-8}
=-30
$