(3/4)(8x-12)=4x-4

Solution for (3/4)(8x-12)=4x-4 equation:

(3/4)(8x-12)=4x-4

We move all terms to the left:

(3/4)(8x-12)-(4x-4)=0


Domain of the equation: 4)(8x-12)!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)(8x-12)-(4x-4)=0

We get rid of parentheses

(+3/4)(8x-12)-4x+4=0

We multiply parentheses ..

(+24x^2+3/4*-12)-4x+4=0

We multiply all the terms by the denominator


(+24x^2+3-4x*4*-12)+4*4*-12)=0

We add all the numbers together, and all the variables

(+24x^2+3-4x*4*-12)=0

We get rid of parentheses

24x^2-4x*4*+3-12=0

We add all the numbers together, and all the variables

24x^2-4x*4*-9=0

Wy multiply elements

24x^2-16x^2-9=0

We add all the numbers together, and all the variables

8x^2-9=0

a = 8; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·8·(-9)
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*8}=\frac{0-12\sqrt{2}}{16}
=-\frac{12\sqrt{2}}{16}
=-\frac{3\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*8}=\frac{0+12\sqrt{2}}{16}
=\frac{12\sqrt{2}}{16}
=\frac{3\sqrt{2}}{4}
$