(4x^2+8x+3)/(4x^2-1)=0

Solution for (4x^2+8x+3)/(4x^2-1)=0 equation:

(4x^2+8x+3)/(4x^2-1)=0


Domain of the equation: (4x^2-1)!=0

We move all terms containing x to the left, all other terms to the right

4x^2!=1

x^2!=1/4

x^2!=√1/4

x!=1

x∈R


We multiply all the terms by the denominator


(4x^2+8x+3)=0

We get rid of parentheses

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
$