1500=(40+2x)(20+2x)

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Solution for 1500=(40+2x)(20+2x) equation: 



1500=(40+2x)(20+2x)

We move all terms to the left:

1500-((40+2x)(20+2x))=0

We add all the numbers together, and all the variables

-((2x+40)(2x+20))+1500=0

We multiply parentheses ..

-((+4x^2+40x+80x+800))+1500=0



We calculate terms in parentheses: -((+4x^2+40x+80x+800)), so:

(+4x^2+40x+80x+800)

We get rid of parentheses

4x^2+40x+80x+800

We add all the numbers together, and all the variables

4x^2+120x+800

Back to the equation:

-(4x^2+120x+800)



We get rid of parentheses

-4x^2-120x-800+1500=0

We add all the numbers together, and all the variables

-4x^2-120x+700=0

a = -4; b = -120; c = +700;
Δ = b2-4ac
Δ = -1202-4·(-4)·700
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-160}{2*-4}=\frac{-40}{-8}
=+5
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+160}{2*-4}=\frac{280}{-8}
=-35
$

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