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

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

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

We move all terms to the left:

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


Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

3/4x+1x-4+6=0

We multiply all the terms by the denominator


1x*4x-4*4x+6*4x+3=0

Wy multiply elements

4x^2-16x+24x+3=0

We add all the numbers together, and all the variables

4x^2+8x+3=0

a = 4; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·4·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*4}=\frac{-12}{8}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*4}=\frac{-4}{8}
=-1/2
$