(3x-4/6)=1/4(2x+3)

Solution for (3x-4/6)=1/4(2x+3) equation:

(3x-4/6)=1/4(2x+3)

We move all terms to the left:

(3x-4/6)-(1/4(2x+3))=0


Domain of the equation: 4(2x+3))!=0

x∈R


We add all the numbers together, and all the variables

(+3x-4/6)-(1/4(2x+3))=0

We get rid of parentheses

3x-(1/4(2x+3))-4/6=0

We calculate fractions


3x+()/48x+(-16x2/48x=0

We multiply all the terms by the denominator


3x*48x+(-16x2+()=0


We calculate terms in parentheses: +(-16x2+(), so:

-16x2+(

We add all the numbers together, and all the variables

-16x^2

Back to the equation:

+(-16x^2)


Wy multiply elements

(-16x^2)+144x^2=0

We get rid of parentheses

-16x^2+144x^2=0

We add all the numbers together, and all the variables

128x^2=0

a = 128; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·128·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{256}=0$