(16+2x)(32+2x)=924

Solution for (16+2x)(32+2x)=924 equation:

(16+2x)(32+2x)=924

We move all terms to the left:

(16+2x)(32+2x)-(924)=0

We add all the numbers together, and all the variables

(2x+16)(2x+32)-924=0

We multiply parentheses ..

(+4x^2+64x+32x+512)-924=0

We get rid of parentheses

4x^2+64x+32x+512-924=0

We add all the numbers together, and all the variables

4x^2+96x-412=0

a = 4; b = 96; c = -412;
Δ = b2-4ac
Δ = 962-4·4·(-412)
Δ = 15808
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:

$\sqrt{\Delta}=\sqrt{15808}=\sqrt{64*247}=\sqrt{64}*\sqrt{247}=8\sqrt{247}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-8\sqrt{247}}{2*4}=\frac{-96-8\sqrt{247}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+8\sqrt{247}}{2*4}=\frac{-96+8\sqrt{247}}{8}
$