3x=1/4x^+5

Solution for 3x=1/4x^+5 equation:

3x=1/4x^+5

We move all terms to the left:

3x-(1/4x^+5)=0


Domain of the equation: 4x^+5)!=0

x∈R


We get rid of parentheses

3x-1/4x^-5=0

We multiply all the terms by the denominator


3x*4x^-5*4x^-1=0

Wy multiply elements

12x^2-20x-1=0

a = 12; b = -20; c = -1;
Δ = b2-4ac
Δ = -202-4·12·(-1)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{7}}{2*12}=\frac{20-8\sqrt{7}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{7}}{2*12}=\frac{20+8\sqrt{7}}{24}
$