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

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

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


Domain of the equation: 4x^2!=0

x^2!=0/4

x^2!=√0

x!=0

x∈R


We multiply parentheses ..

(+2x^2+6x+x+3)/4x^2-1=0

We multiply all the terms by the denominator


(+2x^2+6x+x+3)-1*4x^2=0

Wy multiply elements

(+2x^2+6x+x+3)-4x^2=0

We get rid of parentheses

2x^2-4x^2+6x+x+3=0

We add all the numbers together, and all the variables

-2x^2+7x+3=0

a = -2; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·(-2)·3
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{73}}{2*-2}=\frac{-7-\sqrt{73}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{73}}{2*-2}=\frac{-7+\sqrt{73}}{-4}
$