1/5(20x+40)-2=-1/3(21x-6)

Solution for 1/5(20x+40)-2=-1/3(21x-6) equation:

1/5(20x+40)-2=-1/3(21x-6)

We move all terms to the left:

1/5(20x+40)-2-(-1/3(21x-6))=0


Domain of the equation: 5(20x+40)!=0

x∈R



Domain of the equation: 3(21x-6))!=0

x∈R


We calculate fractions


(3x2/(5(20x+40)*3(21x-6)))+(-(-5x2)/(5(20x+40)*3(21x-6)))-2=0


We calculate terms in parentheses: +(3x2/(5(20x+40)*3(21x-6))), so:

3x2/(5(20x+40)*3(21x-6))

We multiply all the terms by the denominator


3x2

We add all the numbers together, and all the variables

3x^2

Back to the equation:

+(3x^2)



We calculate terms in parentheses: +(-(-5x2)/(5(20x+40)*3(21x-6))), so:

-(-5x2)/(5(20x+40)*3(21x-6))

We add all the numbers together, and all the variables

-(-5x^2)/(5(20x+40)*3(21x-6))

We multiply all the terms by the denominator


-(-5x^2)

We get rid of parentheses

5x^2

Back to the equation:

+(5x^2)


We add all the numbers together, and all the variables

8x^2-2=0

a = 8; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·8·(-2)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*8}=\frac{-8}{16}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*8}=\frac{8}{16}
=1/2
$