(X+9)(x+9)+x+x=302

Solution for (X+9)(x+9)+x+x=302 equation:

(X+9)(X+9)+X+X=302

We move all terms to the left:

(X+9)(X+9)+X+X-(302)=0

We add all the numbers together, and all the variables

2X+(X+9)(X+9)-302=0

We multiply parentheses ..

(+X^2+9X+9X+81)+2X-302=0

We get rid of parentheses

X^2+9X+9X+2X+81-302=0

We add all the numbers together, and all the variables

X^2+20X-221=0

a = 1; b = 20; c = -221;
Δ = b2-4ac
Δ = 202-4·1·(-221)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{321}}{2*1}=\frac{-20-2\sqrt{321}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{321}}{2*1}=\frac{-20+2\sqrt{321}}{2}
$