2x+(3/4x)=33

Solution for 2x+(3/4x)=33 equation:

2x+(3/4x)=33

We move all terms to the left:

2x+(3/4x)-(33)=0


Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2x+(+3/4x)-33=0

We get rid of parentheses

2x+3/4x-33=0

We multiply all the terms by the denominator


2x*4x-33*4x+3=0

Wy multiply elements

8x^2-132x+3=0

a = 8; b = -132; c = +3;
Δ = b2-4ac
Δ = -1322-4·8·3
Δ = 17328
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:

$\sqrt{\Delta}=\sqrt{17328}=\sqrt{5776*3}=\sqrt{5776}*\sqrt{3}=76\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-132)-76\sqrt{3}}{2*8}=\frac{132-76\sqrt{3}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-132)+76\sqrt{3}}{2*8}=\frac{132+76\sqrt{3}}{16}
$