(1/3)x+(1/2)=5(6x+4)/6

Solution for (1/3)x+(1/2)=5(6x+4)/6 equation:

(1/3)x+(1/2)=5(6x+4)/6

We move all terms to the left:

(1/3)x+(1/2)-(5(6x+4)/6)=0


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/3)x-(5(6x+4)/6)+(+1/2)=0

We multiply parentheses

x^2-(5(6x+4)/6)+(+1/2)=0

We get rid of parentheses

x^2-(5(6x+4)/6)+1/2=0

We calculate fractions


x^2+(-(5(6x+4)*2)/()+()/()=0


We calculate terms in parentheses: +(-(5(6x+4)*2)/()+()/(), so:

-(5(6x+4)*2)/()+()/(

We add all the numbers together, and all the variables

-(5(6x+4)*2)/()+1

We multiply all the terms by the denominator


-(5(6x+4)*2)+1*()


We calculate terms in parentheses: -(5(6x+4)*2), so:

5(6x+4)*2

We multiply parentheses

60x+40

Back to the equation:

-(60x+40)


We add all the numbers together, and all the variables

-(60x+40)

We get rid of parentheses

-60x-40

Back to the equation:

+(-60x-40)


We get rid of parentheses

x^2-60x-40=0

a = 1; b = -60; c = -40;
Δ = b2-4ac
Δ = -602-4·1·(-40)
Δ = 3760
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:

$\sqrt{\Delta}=\sqrt{3760}=\sqrt{16*235}=\sqrt{16}*\sqrt{235}=4\sqrt{235}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{235}}{2*1}=\frac{60-4\sqrt{235}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{235}}{2*1}=\frac{60+4\sqrt{235}}{2}
$