(X+2)/2=(2x+5)/3-x+4

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

(X+2)/2=(2X+5)/3-X+4

We move all terms to the left:

(X+2)/2-((2X+5)/3-X+4)=0


Domain of the equation: 3-X+4)!=0

We move all terms containing X to the left, all other terms to the right

-X+4)!=-3

X∈R


We calculate fractions


(2X+6)/(-X)+(-((2X+5)*2)/(-X)=0


We calculate terms in parentheses: +(-((2X+5)*2)/(-X), so:

-((2X+5)*2)/(-X

We multiply all the terms by the denominator


-((2X+5)*2)


We calculate terms in parentheses: -((2X+5)*2), so:

(2X+5)*2

We multiply parentheses

4X+10

Back to the equation:

-(4X+10)


We get rid of parentheses

-4X-10

Back to the equation:

+(-4X-10)


We add all the numbers together, and all the variables

(2X+6)/(-1X)+(-4X-10)=0

We get rid of parentheses

(2X+6)/(-1X)-4X-10=0

We multiply all the terms by the denominator


(2X+6)-4X*(-1X)-10*(-1X)=0

We multiply parentheses

4X^2+(2X+6)+10X=0

We get rid of parentheses

4X^2+2X+10X+6=0

We add all the numbers together, and all the variables

4X^2+12X+6=0

a = 4; b = 12; c = +6;
Δ = b2-4ac
Δ = 122-4·4·6
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{-(12)-4\sqrt{3}}{2*4}=\frac{-12-4\sqrt{3}}{8}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{3}}{2*4}=\frac{-12+4\sqrt{3}}{8}
$