(7/3)(3x+2)=5

Solution for (7/3)(3x+2)=5 equation:

(7/3)(3x+2)=5

We move all terms to the left:

(7/3)(3x+2)-(5)=0


Domain of the equation: 3)(3x+2)!=0

x∈R


We add all the numbers together, and all the variables

(+7/3)(3x+2)-5=0

We multiply parentheses ..

(+21x^2+7/3*2)-5=0

We multiply all the terms by the denominator


(+21x^2+7-5*3*2)=0

We get rid of parentheses

21x^2+7-5*3*2=0

We add all the numbers together, and all the variables

21x^2-23=0

a = 21; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·21·(-23)
Δ = 1932
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:

$\sqrt{\Delta}=\sqrt{1932}=\sqrt{4*483}=\sqrt{4}*\sqrt{483}=2\sqrt{483}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{483}}{2*21}=\frac{0-2\sqrt{483}}{42}
=-\frac{2\sqrt{483}}{42}
=-\frac{\sqrt{483}}{21}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{483}}{2*21}=\frac{0+2\sqrt{483}}{42}
=\frac{2\sqrt{483}}{42}
=\frac{\sqrt{483}}{21}
$