(6r^2-19r-7)/(2r-7)=4r-3

Solution for (6r^2-19r-7)/(2r-7)=4r-3 equation:

(6r^2-19r-7)/(2r-7)=4r-3

We move all terms to the left:

(6r^2-19r-7)/(2r-7)-(4r-3)=0


Domain of the equation: (2r-7)!=0

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

2r!=7

r!=7/2

r!=3+1/2

r∈R


We get rid of parentheses

(6r^2-19r-7)/(2r-7)-4r+3=0

We multiply all the terms by the denominator


(6r^2-19r-7)-4r*(2r-7)+3*(2r-7)=0

We multiply parentheses

-8r^2+(6r^2-19r-7)+28r+6r-21=0

We get rid of parentheses

-8r^2+6r^2-19r+28r+6r-7-21=0

We add all the numbers together, and all the variables

-2r^2+15r-28=0

a = -2; b = 15; c = -28;
Δ = b2-4ac
Δ = 152-4·(-2)·(-28)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-1}{2*-2}=\frac{-16}{-4}
=+4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+1}{2*-2}=\frac{-14}{-4}
=3+1/2
$