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^2-7x+10=0

a = 1; b = -7; c = +10;
Δ = b2-4ac
Δ = -72-4·1·10
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*1}=\frac{10}{2}
=5
$