(17+2w)(13+2w)+221=396

Solution for (17+2w)(13+2w)+221=396 equation:

(17+2w)(13+2w)+221=396

We move all terms to the left:

(17+2w)(13+2w)+221-(396)=0

We add all the numbers together, and all the variables

(2w+17)(2w+13)+221-396=0

We add all the numbers together, and all the variables

(2w+17)(2w+13)-175=0

We multiply parentheses ..

(+4w^2+26w+34w+221)-175=0

We get rid of parentheses

4w^2+26w+34w+221-175=0

We add all the numbers together, and all the variables

4w^2+60w+46=0

a = 4; b = 60; c = +46;
Δ = b2-4ac
Δ = 602-4·4·46
Δ = 2864
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:

$\sqrt{\Delta}=\sqrt{2864}=\sqrt{16*179}=\sqrt{16}*\sqrt{179}=4\sqrt{179}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-4\sqrt{179}}{2*4}=\frac{-60-4\sqrt{179}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+4\sqrt{179}}{2*4}=\frac{-60+4\sqrt{179}}{8}
$