(x+4)(2x+4)=196

Solution for (x+4)(2x+4)=196 equation:

(x+4)(2x+4)=196

We move all terms to the left:

(x+4)(2x+4)-(196)=0

We multiply parentheses ..

(+2x^2+4x+8x+16)-196=0

We get rid of parentheses

2x^2+4x+8x+16-196=0

We add all the numbers together, and all the variables

2x^2+12x-180=0

a = 2; b = 12; c = -180;
Δ = b2-4ac
Δ = 122-4·2·(-180)
Δ = 1584
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:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{11}}{2*2}=\frac{-12-12\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{11}}{2*2}=\frac{-12+12\sqrt{11}}{4}
$