(x+5)+(x+5)+(2x+4)=4x(2x+3)

Solution for (x+5)+(x+5)+(2x+4)=4x(2x+3) equation:

(x+5)+(x+5)+(2x+4)=4x(2x+3)

We move all terms to the left:

(x+5)+(x+5)+(2x+4)-(4x(2x+3))=0

We get rid of parentheses

x+x+2x-(4x(2x+3))+5+5+4=0


We calculate terms in parentheses: -(4x(2x+3)), so:

4x(2x+3)

We multiply parentheses

8x^2+12x

Back to the equation:

-(8x^2+12x)


We add all the numbers together, and all the variables

4x-(8x^2+12x)+14=0

We get rid of parentheses

-8x^2+4x-12x+14=0

We add all the numbers together, and all the variables

-8x^2-8x+14=0

a = -8; b = -8; c = +14;
Δ = b2-4ac
Δ = -82-4·(-8)·14
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16\sqrt{2}}{2*-8}=\frac{8-16\sqrt{2}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16\sqrt{2}}{2*-8}=\frac{8+16\sqrt{2}}{-16}
$