40=(10+2x)(8+2x)

Solution for 40=(10+2x)(8+2x) equation:

40=(10+2x)(8+2x)

We move all terms to the left:

40-((10+2x)(8+2x))=0

We add all the numbers together, and all the variables

-((2x+10)(2x+8))+40=0

We multiply parentheses ..

-((+4x^2+16x+20x+80))+40=0


We calculate terms in parentheses: -((+4x^2+16x+20x+80)), so:

(+4x^2+16x+20x+80)

We get rid of parentheses

4x^2+16x+20x+80

We add all the numbers together, and all the variables

4x^2+36x+80

Back to the equation:

-(4x^2+36x+80)


We get rid of parentheses

-4x^2-36x-80+40=0

We add all the numbers together, and all the variables

-4x^2-36x-40=0

a = -4; b = -36; c = -40;
Δ = b2-4ac
Δ = -362-4·(-4)·(-40)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{41}}{2*-4}=\frac{36-4\sqrt{41}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{41}}{2*-4}=\frac{36+4\sqrt{41}}{-8}
$