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

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

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

We move all terms to the left:

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


Domain of the equation: 5)(x+2)!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

(+3x^2+3-x*5*2)=0

We get rid of parentheses

3x^2-x*5*2+3=0

Wy multiply elements

3x^2-10x*2+3=0

Wy multiply elements

3x^2-20x+3=0

a = 3; b = -20; c = +3;
Δ = b2-4ac
Δ = -202-4·3·3
Δ = 364
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:

$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{91}}{2*3}=\frac{20-2\sqrt{91}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{91}}{2*3}=\frac{20+2\sqrt{91}}{6}
$