(6x^2+13x+5)/(2x+1)=0

Solution for (6x^2+13x+5)/(2x+1)=0 equation:

(6x^2+13x+5)/(2x+1)=0


Domain of the equation: (2x+1)!=0

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

2x!=-1

x!=-1/2

x!=-1/2

x∈R


We multiply all the terms by the denominator


(6x^2+13x+5)=0

We get rid of parentheses

6x^2+13x+5=0

a = 6; b = 13; c = +5;
Δ = b2-4ac
Δ = 132-4·6·5
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-7}{2*6}=\frac{-20}{12}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+7}{2*6}=\frac{-6}{12}
=-1/2
$