4x(x+3)=3x+4+x

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

4x(x+3)=3x+4+x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2+8x-4=0

a = 4; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·4·(-4)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*4}=\frac{-8-8\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*4}=\frac{-8+8\sqrt{2}}{8}
$