(16+2x)(8+2x)-128=200

Solution for (16+2x)(8+2x)-128=200 equation:

(16+2x)(8+2x)-128=200

We move all terms to the left:

(16+2x)(8+2x)-128-(200)=0

We add all the numbers together, and all the variables

(2x+16)(2x+8)-128-200=0

We add all the numbers together, and all the variables

(2x+16)(2x+8)-328=0

We multiply parentheses ..

(+4x^2+16x+32x+128)-328=0

We get rid of parentheses

4x^2+16x+32x+128-328=0

We add all the numbers together, and all the variables

4x^2+48x-200=0

a = 4; b = 48; c = -200;
Δ = b2-4ac
Δ = 482-4·4·(-200)
Δ = 5504
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:

$\sqrt{\Delta}=\sqrt{5504}=\sqrt{64*86}=\sqrt{64}*\sqrt{86}=8\sqrt{86}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{86}}{2*4}=\frac{-48-8\sqrt{86}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{86}}{2*4}=\frac{-48+8\sqrt{86}}{8}
$