12^2=(8+x+5)(x+5)

Solution for 12^2=(8+x+5)(x+5) equation:

12^2=(8+x+5)(x+5)

We move all terms to the left:

12^2-((8+x+5)(x+5))=0

We add all the numbers together, and all the variables

-((x+13)(x+5))+12^2=0

We add all the numbers together, and all the variables

-((x+13)(x+5))+144=0

We multiply parentheses ..

-((+x^2+5x+13x+65))+144=0


We calculate terms in parentheses: -((+x^2+5x+13x+65)), so:

(+x^2+5x+13x+65)

We get rid of parentheses

x^2+5x+13x+65

We add all the numbers together, and all the variables

x^2+18x+65

Back to the equation:

-(x^2+18x+65)


We get rid of parentheses

-x^2-18x-65+144=0

We add all the numbers together, and all the variables

-1x^2-18x+79=0

a = -1; b = -18; c = +79;
Δ = b2-4ac
Δ = -182-4·(-1)·79
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-8\sqrt{10}}{2*-1}=\frac{18-8\sqrt{10}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+8\sqrt{10}}{2*-1}=\frac{18+8\sqrt{10}}{-2}
$