(2/3)(9x+1)+16=18

Solution for (2/3)(9x+1)+16=18 equation:

(2/3)(9x+1)+16=18

We move all terms to the left:

(2/3)(9x+1)+16-(18)=0


Domain of the equation: 3)(9x+1)!=0

x∈R


We add all the numbers together, and all the variables

(+2/3)(9x+1)+16-18=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+18x^2+2/3*1)-2=0

We multiply all the terms by the denominator


(+18x^2+2-2*3*1)=0

We get rid of parentheses

18x^2+2-2*3*1=0

We add all the numbers together, and all the variables

18x^2-4=0

a = 18; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·18·(-4)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*18}=\frac{0-12\sqrt{2}}{36}
=-\frac{12\sqrt{2}}{36}
=-\frac{\sqrt{2}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*18}=\frac{0+12\sqrt{2}}{36}
=\frac{12\sqrt{2}}{36}
=\frac{\sqrt{2}}{3}
$