2x^2+2X+X=4-X

Solution for 2x^2+2X+X=4-X equation:

2x^2+2x+x=4-x

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+2x+x-(-1x+4)=0

We add all the numbers together, and all the variables

2x^2+3x-(-1x+4)=0

We get rid of parentheses

2x^2+3x+1x-4=0

We add all the numbers together, and all the variables

2x^2+4x-4=0

a = 2; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·2·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*2}=\frac{-4-4\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*2}=\frac{-4+4\sqrt{3}}{4}
$