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

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

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

We move all terms to the left:

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

We multiply parentheses ..

2x^2-((+x^2-2x-4x+8)+2x)-3x+2=0


We calculate terms in parentheses: -((+x^2-2x-4x+8)+2x), so:

(+x^2-2x-4x+8)+2x

We get rid of parentheses

x^2-2x-4x+2x+8

We add all the numbers together, and all the variables

x^2-4x+8

Back to the equation:

-(x^2-4x+8)


We add all the numbers together, and all the variables

2x^2-3x-(x^2-4x+8)+2=0

We get rid of parentheses

2x^2-x^2-3x+4x-8+2=0

We add all the numbers together, and all the variables

x^2+x-6=0

a = 1; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*1}=\frac{4}{2}
=2
$