(3/4)(2x+12)=3x-3

Solution for (3/4)(2x+12)=3x-3 equation:

(3/4)(2x+12)=3x-3

We move all terms to the left:

(3/4)(2x+12)-(3x-3)=0


Domain of the equation: 4)(2x+12)!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)(2x+12)-(3x-3)=0

We get rid of parentheses

(+3/4)(2x+12)-3x+3=0

We multiply parentheses ..

(+6x^2+3/4*12)-3x+3=0

We multiply all the terms by the denominator


(+6x^2+3-3x*4*12)+3*4*12)=0

We add all the numbers together, and all the variables

(+6x^2+3-3x*4*12)=0

We get rid of parentheses

6x^2-3x*4*12+3=0

Wy multiply elements

6x^2-144x*1+3=0

Wy multiply elements

6x^2-144x+3=0

a = 6; b = -144; c = +3;
Δ = b2-4ac
Δ = -1442-4·6·3
Δ = 20664
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:

$\sqrt{\Delta}=\sqrt{20664}=\sqrt{36*574}=\sqrt{36}*\sqrt{574}=6\sqrt{574}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-144)-6\sqrt{574}}{2*6}=\frac{144-6\sqrt{574}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-144)+6\sqrt{574}}{2*6}=\frac{144+6\sqrt{574}}{12}
$