(12x^2+4x-3)/3x=0

Solution for (12x^2+4x-3)/3x=0 equation:

(12x^2+4x-3)/3x=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We multiply all the terms by the denominator


(12x^2+4x-3)=0

We get rid of parentheses

12x^2+4x-3=0

a = 12; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·12·(-3)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*12}=\frac{-4-4\sqrt{10}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*12}=\frac{-4+4\sqrt{10}}{24}
$