1/3(21x+12)-20=-1/2(10x-18)

Solution for 1/3(21x+12)-20=-1/2(10x-18) equation:

1/3(21x+12)-20=-1/2(10x-18)

We move all terms to the left:

1/3(21x+12)-20-(-1/2(10x-18))=0


Domain of the equation: 3(21x+12)!=0

x∈R



Domain of the equation: 2(10x-18))!=0

x∈R


We calculate fractions


(2x1/(3(21x+12)*2(10x-18)))+(-(-3x2)/(3(21x+12)*2(10x-18)))-20=0


We calculate terms in parentheses: +(2x1/(3(21x+12)*2(10x-18))), so:

2x1/(3(21x+12)*2(10x-18))

We multiply all the terms by the denominator


2x1

We add all the numbers together, and all the variables

2x

Back to the equation:

+(2x)



We calculate terms in parentheses: +(-(-3x2)/(3(21x+12)*2(10x-18))), so:

-(-3x2)/(3(21x+12)*2(10x-18))

We add all the numbers together, and all the variables

-(-3x^2)/(3(21x+12)*2(10x-18))

We multiply all the terms by the denominator


-(-3x^2)

We get rid of parentheses

3x^2

Back to the equation:

+(3x^2)


determiningTheFunctionDomain
3x^2+2x-20=0

a = 3; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·3·(-20)
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{61}}{2*3}=\frac{-2-2\sqrt{61}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{61}}{2*3}=\frac{-2+2\sqrt{61}}{6}
$