48=(6+2x)(4+2x)

Solution for 48=(6+2x)(4+2x) equation:

48=(6+2x)(4+2x)

We move all terms to the left:

48-((6+2x)(4+2x))=0

We add all the numbers together, and all the variables

-((2x+6)(2x+4))+48=0

We multiply parentheses ..

-((+4x^2+8x+12x+24))+48=0


We calculate terms in parentheses: -((+4x^2+8x+12x+24)), so:

(+4x^2+8x+12x+24)

We get rid of parentheses

4x^2+8x+12x+24

We add all the numbers together, and all the variables

4x^2+20x+24

Back to the equation:

-(4x^2+20x+24)


We get rid of parentheses

-4x^2-20x-24+48=0

We add all the numbers together, and all the variables

-4x^2-20x+24=0

a = -4; b = -20; c = +24;
Δ = b2-4ac
Δ = -202-4·(-4)·24
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-28}{2*-4}=\frac{-8}{-8}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+28}{2*-4}=\frac{48}{-8}
=-6
$