(-8x^2-6x+5)/(2x+1)=0

Solution for (-8x^2-6x+5)/(2x+1)=0 equation:

(-8x^2-6x+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


(-8x^2-6x+5)=0

We get rid of parentheses

-8x^2-6x+5=0

a = -8; b = -6; c = +5;
Δ = b2-4ac
Δ = -62-4·(-8)·5
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-14}{2*-8}=\frac{-8}{-16}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+14}{2*-8}=\frac{20}{-16}
=-1+1/4
$