2x(12+x)=(x+4)2

Solution for 2x(12+x)=(x+4)2 equation:

2x(12+x)=(x+4)2

We move all terms to the left:

2x(12+x)-((x+4)2)=0

We add all the numbers together, and all the variables

2x(x+12)-((x+4)2)=0

We multiply parentheses

2x^2+24x-((x+4)2)=0


We calculate terms in parentheses: -((x+4)2), so:

(x+4)2

We multiply parentheses

2x+8

Back to the equation:

-(2x+8)


We get rid of parentheses

2x^2+24x-2x-8=0

We add all the numbers together, and all the variables

2x^2+22x-8=0

a = 2; b = 22; c = -8;
Δ = b2-4ac
Δ = 222-4·2·(-8)
Δ = 548
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:

$\sqrt{\Delta}=\sqrt{548}=\sqrt{4*137}=\sqrt{4}*\sqrt{137}=2\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{137}}{2*2}=\frac{-22-2\sqrt{137}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{137}}{2*2}=\frac{-22+2\sqrt{137}}{4}
$