16-3x=2/3x+5

Solution for 16-3x=2/3x+5 equation:

16-3x=2/3x+5

We move all terms to the left:

16-3x-(2/3x+5)=0


Domain of the equation: 3x+5)!=0

x∈R


We get rid of parentheses

-3x-2/3x-5+16=0

We multiply all the terms by the denominator


-3x*3x-5*3x+16*3x-2=0

Wy multiply elements

-9x^2-15x+48x-2=0

We add all the numbers together, and all the variables

-9x^2+33x-2=0

a = -9; b = 33; c = -2;
Δ = b2-4ac
Δ = 332-4·(-9)·(-2)
Δ = 1017
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:

$\sqrt{\Delta}=\sqrt{1017}=\sqrt{9*113}=\sqrt{9}*\sqrt{113}=3\sqrt{113}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{113}}{2*-9}=\frac{-33-3\sqrt{113}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{113}}{2*-9}=\frac{-33+3\sqrt{113}}{-18}
$