63=(7+2x)(5+2x)

Solution for 63=(7+2x)(5+2x) equation:

63=(7+2x)(5+2x)

We move all terms to the left:

63-((7+2x)(5+2x))=0

We add all the numbers together, and all the variables

-((2x+7)(2x+5))+63=0

We multiply parentheses ..

-((+4x^2+10x+14x+35))+63=0


We calculate terms in parentheses: -((+4x^2+10x+14x+35)), so:

(+4x^2+10x+14x+35)

We get rid of parentheses

4x^2+10x+14x+35

We add all the numbers together, and all the variables

4x^2+24x+35

Back to the equation:

-(4x^2+24x+35)


We get rid of parentheses

-4x^2-24x-35+63=0

We add all the numbers together, and all the variables

-4x^2-24x+28=0

a = -4; b = -24; c = +28;
Δ = b2-4ac
Δ = -242-4·(-4)·28
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-32}{2*-4}=\frac{-8}{-8}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+32}{2*-4}=\frac{56}{-8}
=-7
$