(-8x^2-6x+4)/(2x+1)=0

Solution for (-8x^2-6x+4)/(2x+1)=0 equation:

(-8x^2-6x+4)/(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


(-8x^2-6x+4)=0

We get rid of parentheses

-8x^2-6x+4=0

a = -8; b = -6; c = +4;
Δ = b2-4ac
Δ = -62-4·(-8)·4
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{41}}{2*-8}=\frac{6-2\sqrt{41}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{41}}{2*-8}=\frac{6+2\sqrt{41}}{-16}
$