(11+2x)(3+2x)=308

Solution for (11+2x)(3+2x)=308 equation:

(11+2x)(3+2x)=308

We move all terms to the left:

(11+2x)(3+2x)-(308)=0

We add all the numbers together, and all the variables

(2x+11)(2x+3)-308=0

We multiply parentheses ..

(+4x^2+6x+22x+33)-308=0

We get rid of parentheses

4x^2+6x+22x+33-308=0

We add all the numbers together, and all the variables

4x^2+28x-275=0

a = 4; b = 28; c = -275;
Δ = b2-4ac
Δ = 282-4·4·(-275)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-72}{2*4}=\frac{-100}{8}
=-12+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+72}{2*4}=\frac{44}{8}
=5+1/2
$