(16+2w)(14+2w)=396

Solution for (16+2w)(14+2w)=396 equation:

(16+2w)(14+2w)=396

We move all terms to the left:

(16+2w)(14+2w)-(396)=0

We add all the numbers together, and all the variables

(2w+16)(2w+14)-396=0

We multiply parentheses ..

(+4w^2+28w+32w+224)-396=0

We get rid of parentheses

4w^2+28w+32w+224-396=0

We add all the numbers together, and all the variables

4w^2+60w-172=0

a = 4; b = 60; c = -172;
Δ = b2-4ac
Δ = 602-4·4·(-172)
Δ = 6352
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6352}=\sqrt{16*397}=\sqrt{16}*\sqrt{397}=4\sqrt{397}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-4\sqrt{397}}{2*4}=\frac{-60-4\sqrt{397}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+4\sqrt{397}}{2*4}=\frac{-60+4\sqrt{397}}{8}
$