99=(7+2x)(5+2x)

Solution for 99=(7+2x)(5+2x) equation:

99=(7+2x)(5+2x)

We move all terms to the left:

99-((7+2x)(5+2x))=0

We add all the numbers together, and all the variables

-((2x+7)(2x+5))+99=0

We multiply parentheses ..

-((+4x^2+10x+14x+35))+99=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+99=0

We add all the numbers together, and all the variables

-4x^2-24x+64=0

a = -4; b = -24; c = +64;
Δ = b2-4ac
Δ = -242-4·(-4)·64
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-40}{2*-4}=\frac{-16}{-8}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+40}{2*-4}=\frac{64}{-8}
=-8
$