2x2-4x+5/3=(x1)(x+2)/2

Solution for 2x2-4x+5/3=(x1)(x+2)/2 equation:

2x^2-4x+5/3=(x1)(x+2)/2

We move all terms to the left:

2x^2-4x+5/3-((x1)(x+2)/2)=0

We calculate fractions


2x^2-4x+(-(x1(x+2)*3)/()+()/()=0


We calculate terms in parentheses: +(-(x1(x+2)*3)/()+()/(), so:

-(x1(x+2)*3)/()+()/(

We add all the numbers together, and all the variables

-(x1(x+2)*3)/()+1

We multiply all the terms by the denominator


-(x1(x+2)*3)+1*()


We calculate terms in parentheses: -(x1(x+2)*3), so:

x1(x+2)*3

We multiply parentheses

3x^2+6x

Back to the equation:

-(3x^2+6x)


We add all the numbers together, and all the variables

-(3x^2+6x)

We get rid of parentheses

-3x^2-6x

Back to the equation:

+(-3x^2-6x)


We get rid of parentheses

2x^2-3x^2-6x-4x=0

We add all the numbers together, and all the variables

-1x^2-10x=0

a = -1; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·(-1)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*-1}=\frac{0}{-2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*-1}=\frac{20}{-2}
=-10
$