(8+2x)(6+2x)=99

Solution for (8+2x)(6+2x)=99 equation:

(8+2x)(6+2x)=99

We move all terms to the left:

(8+2x)(6+2x)-(99)=0

We add all the numbers together, and all the variables

(2x+8)(2x+6)-99=0

We multiply parentheses ..

(+4x^2+12x+16x+48)-99=0

We get rid of parentheses

4x^2+12x+16x+48-99=0

We add all the numbers together, and all the variables

4x^2+28x-51=0

a = 4; b = 28; c = -51;
Δ = b2-4ac
Δ = 282-4·4·(-51)
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{-(28)-40}{2*4}=\frac{-68}{8}
=-8+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+40}{2*4}=\frac{12}{8}
=1+1/2
$