(3/4)(x-12)+9(x-12)=39

Solution for (3/4)(x-12)+9(x-12)=39 equation:

(3/4)(x-12)+9(x-12)=39

We move all terms to the left:

(3/4)(x-12)+9(x-12)-(39)=0


Domain of the equation: 4)(x-12)!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)(x-12)+9(x-12)-39=0

We multiply parentheses

(+3/4)(x-12)+9x-108-39=0

We multiply parentheses ..

(+3x^2+3/4*-12)+9x-108-39=0

We multiply all the terms by the denominator


(+3x^2+3+9x*4*-12)-108*4*-12)-39*4*-12)=0

We add all the numbers together, and all the variables

(+3x^2+3+9x*4*-12)=0

We get rid of parentheses

3x^2+9x*4*+3-12=0

We add all the numbers together, and all the variables

3x^2+9x*4*-9=0

Wy multiply elements

3x^2+36x^2-9=0

We add all the numbers together, and all the variables

39x^2-9=0

a = 39; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·39·(-9)
Δ = 1404
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:

$\sqrt{\Delta}=\sqrt{1404}=\sqrt{36*39}=\sqrt{36}*\sqrt{39}=6\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{39}}{2*39}=\frac{0-6\sqrt{39}}{78}
=-\frac{6\sqrt{39}}{78}
=-\frac{\sqrt{39}}{13}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{39}}{2*39}=\frac{0+6\sqrt{39}}{78}
=\frac{6\sqrt{39}}{78}
=\frac{\sqrt{39}}{13}
$