(8r^2-14r+3)/r+5=4r-1

Solution for (8r^2-14r+3)/r+5=4r-1 equation:

(8r^2-14r+3)/r+5=4r-1

We move all terms to the left:

(8r^2-14r+3)/r+5-(4r-1)=0


Domain of the equation: r!=0

r∈R


We get rid of parentheses

(8r^2-14r+3)/r-4r+1+5=0

We multiply all the terms by the denominator


(8r^2-14r+3)-4r*r+1*r+5*r=0

We add all the numbers together, and all the variables

6r+(8r^2-14r+3)-4r*r=0

Wy multiply elements

-4r^2+6r+(8r^2-14r+3)=0

We get rid of parentheses

-4r^2+8r^2+6r-14r+3=0

We add all the numbers together, and all the variables

4r^2-8r+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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*4}=\frac{4}{8}
=1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*4}=\frac{12}{8}
=1+1/2
$