2(2x^2-3x+2)=6+x

Solution for 2(2x^2-3x+2)=6+x equation:

2(2x^2-3x+2)=6+x

We move all terms to the left:

2(2x^2-3x+2)-(6+x)=0

We add all the numbers together, and all the variables

2(2x^2-3x+2)-(x+6)=0

We multiply parentheses

4x^2-6x-(x+6)+4=0

We get rid of parentheses

4x^2-6x-x-6+4=0

We add all the numbers together, and all the variables

4x^2-7x-2=0

a = 4; b = -7; c = -2;
Δ = b2-4ac
Δ = -72-4·4·(-2)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-9}{2*4}=\frac{-2}{8}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+9}{2*4}=\frac{16}{8}
=2
$