(2/3)(2x+3)-12=-6

Solution for (2/3)(2x+3)-12=-6 equation:

(2/3)(2x+3)-12=-6

We move all terms to the left:

(2/3)(2x+3)-12-(-6)=0


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

x∈R


We add all the numbers together, and all the variables

(+2/3)(2x+3)-12-(-6)=0

We add all the numbers together, and all the variables

(+2/3)(2x+3)-6=0

We multiply parentheses ..

(+4x^2+2/3*3)-6=0

We multiply all the terms by the denominator


(+4x^2+2-6*3*3)=0

We get rid of parentheses

4x^2+2-6*3*3=0

We add all the numbers together, and all the variables

4x^2-52=0

a = 4; b = 0; c = -52;
Δ = b2-4ac
Δ = 02-4·4·(-52)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{13}}{2*4}=\frac{0-8\sqrt{13}}{8}
=-\frac{8\sqrt{13}}{8}
=-\sqrt{13}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{13}}{2*4}=\frac{0+8\sqrt{13}}{8}
=\frac{8\sqrt{13}}{8}
=\sqrt{13}
$