2/3x+9=1x+5

Solution for 2/3x+9=1x+5 equation:

2/3x+9=1x+5

We move all terms to the left:

2/3x+9-(1x+5)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

2/3x-(x+5)+9=0

We get rid of parentheses

2/3x-x-5+9=0

We multiply all the terms by the denominator


-x*3x-5*3x+9*3x+2=0

Wy multiply elements

-3x^2-15x+27x+2=0

We add all the numbers together, and all the variables

-3x^2+12x+2=0

a = -3; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·(-3)·2
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{42}}{2*-3}=\frac{-12-2\sqrt{42}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{42}}{2*-3}=\frac{-12+2\sqrt{42}}{-6}
$