(8+2x)(8+2x)=(x^2+48)

Solution for (8+2x)(8+2x)=(x^2+48) equation:

(8+2x)(8+2x)=(x^2+48)

We move all terms to the left:

(8+2x)(8+2x)-((x^2+48))=0

We add all the numbers together, and all the variables

(2x+8)(2x+8)-((x^2+48))=0

We multiply parentheses ..

(+4x^2+16x+16x+64)-((x^2+48))=0


We calculate terms in parentheses: -((x^2+48)), so:

(x^2+48)

We get rid of parentheses

x^2+48

Back to the equation:

-(x^2+48)


We get rid of parentheses

4x^2-x^2+16x+16x+64-48=0

We add all the numbers together, and all the variables

3x^2+32x+16=0

a = 3; b = 32; c = +16;
Δ = b2-4ac
Δ = 322-4·3·16
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{13}}{2*3}=\frac{-32-8\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{13}}{2*3}=\frac{-32+8\sqrt{13}}{6}
$