3/8(16x-64)=1/2(2x+4)

Solution for 3/8(16x-64)=1/2(2x+4) equation:

3/8(16x-64)=1/2(2x+4)

We move all terms to the left:

3/8(16x-64)-(1/2(2x+4))=0


Domain of the equation: 8(16x-64)!=0

x∈R



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

x∈R


We calculate fractions


(6x2/(8(16x-64)*2(2x+4)))+(-8x1/(8(16x-64)*2(2x+4)))=0


We calculate terms in parentheses: +(6x2/(8(16x-64)*2(2x+4))), so:

6x2/(8(16x-64)*2(2x+4))

We multiply all the terms by the denominator


6x2

We add all the numbers together, and all the variables

6x^2

Back to the equation:

+(6x^2)



We calculate terms in parentheses: +(-8x1/(8(16x-64)*2(2x+4))), so:

-8x1/(8(16x-64)*2(2x+4))

We multiply all the terms by the denominator


-8x1

We add all the numbers together, and all the variables

-8x

Back to the equation:

+(-8x)


We get rid of parentheses

6x^2-8x=0

a = 6; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·6·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*6}=\frac{0}{12}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*6}=\frac{16}{12}
=1+1/3
$