30-(1/24)(3x)=20-(1-24)(2x)

Solution for 30-(1/24)(3x)=20-(1-24)(2x) equation:

30-(1/24)(3x)=20-(1-24)(2x)

We move all terms to the left:

30-(1/24)(3x)-(20-(1-24)(2x))=0


Domain of the equation: 24)3x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

-(+1/24)3x-(20-(-23)2x)+30=0

We multiply parentheses

-3x^2-(20-(-23)2x)+30=0


We calculate terms in parentheses: -(20-(-23)2x), so:

20-(-23)2x

determiningTheFunctionDomain
-(-23)2x+20

We multiply parentheses

46x+20

Back to the equation:

-(46x+20)


We get rid of parentheses

-3x^2-46x-20+30=0

We add all the numbers together, and all the variables

-3x^2-46x+10=0

a = -3; b = -46; c = +10;
Δ = b2-4ac
Δ = -462-4·(-3)·10
Δ = 2236
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:

$\sqrt{\Delta}=\sqrt{2236}=\sqrt{4*559}=\sqrt{4}*\sqrt{559}=2\sqrt{559}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-2\sqrt{559}}{2*-3}=\frac{46-2\sqrt{559}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+2\sqrt{559}}{2*-3}=\frac{46+2\sqrt{559}}{-6}
$