(61/3)(4a+1)=(61/2)a

Solution for (61/3)(4a+1)=(61/2)a equation:

(61/3)(4a+1)=(61/2)a

We move all terms to the left:

(61/3)(4a+1)-((61/2)a)=0


Domain of the equation: 3)(4a+1)!=0

a∈R



Domain of the equation: 2)a)!=0

a!=0/1

a!=0

a∈R


We add all the numbers together, and all the variables

(+61/3)(4a+1)-((+61/2)a)=0

We multiply parentheses ..

(+244a^2+61/3*1)-((+61/2)a)=0

We calculate fractions


(244a^2+122a)/6a^2+()/6a^2=0

We multiply all the terms by the denominator


(244a^2+122a)+()=0

We add all the numbers together, and all the variables

(244a^2+122a)=0

We get rid of parentheses

244a^2+122a=0

a = 244; b = 122; c = 0;
Δ = b2-4ac
Δ = 1222-4·244·0
Δ = 14884
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14884}=122$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(122)-122}{2*244}=\frac{-244}{488}
=-1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(122)+122}{2*244}=\frac{0}{488}
=0
$