(16+2x)(12+2x)=477

Solution for (16+2x)(12+2x)=477 equation:

(16+2x)(12+2x)=477

We move all terms to the left:

(16+2x)(12+2x)-(477)=0

We add all the numbers together, and all the variables

(2x+16)(2x+12)-477=0

We multiply parentheses ..

(+4x^2+24x+32x+192)-477=0

We get rid of parentheses

4x^2+24x+32x+192-477=0

We add all the numbers together, and all the variables

4x^2+56x-285=0

a = 4; b = 56; c = -285;
Δ = b2-4ac
Δ = 562-4·4·(-285)
Δ = 7696
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:

$\sqrt{\Delta}=\sqrt{7696}=\sqrt{16*481}=\sqrt{16}*\sqrt{481}=4\sqrt{481}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-4\sqrt{481}}{2*4}=\frac{-56-4\sqrt{481}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+4\sqrt{481}}{2*4}=\frac{-56+4\sqrt{481}}{8}
$