7+3x=4(2+3/4x)

Solution for 7+3x=4(2+3/4x) equation:

7+3x=4(2+3/4x)

We move all terms to the left:

7+3x-(4(2+3/4x))=0


Domain of the equation: 4x))!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

3x-(4(3/4x+2))+7=0

We multiply all the terms by the denominator


3x*4x+7*4x+2))-(4(3+2))=0

We add all the numbers together, and all the variables

3x*4x+7*4x+2))-(45)=0

We add all the numbers together, and all the variables

3x*4x+7*4x=0

Wy multiply elements

12x^2+28x=0

a = 12; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·12·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*12}=\frac{-56}{24}
=-2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*12}=\frac{0}{24}
=0
$