4x-2-(3/4)(8x-12)=13

Solution for 4x-2-(3/4)(8x-12)=13 equation:

4x-2-(3/4)(8x-12)=13

We move all terms to the left:

4x-2-(3/4)(8x-12)-(13)=0


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

x∈R


We add all the numbers together, and all the variables

4x-(+3/4)(8x-12)-2-13=0

We add all the numbers together, and all the variables

4x-(+3/4)(8x-12)-15=0

We multiply parentheses ..

-(+24x^2+3/4*-12)+4x-15=0

We multiply all the terms by the denominator


-(+24x^2+3+4x*4*-12)-15*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

-40x^2+9=0

a = -40; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-40)·9
Δ = 1440
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:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{10}}{2*-40}=\frac{0-12\sqrt{10}}{-80}
=-\frac{12\sqrt{10}}{-80}
=-\frac{3\sqrt{10}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{10}}{2*-40}=\frac{0+12\sqrt{10}}{-80}
=\frac{12\sqrt{10}}{-80}
=\frac{3\sqrt{10}}{-20}
$