4r+9/4r=5r-3/4r+1-3

Solution for 4r+9/4r=5r-3/4r+1-3 equation:

4r+9/4r=5r-3/4r+1-3

We move all terms to the left:

4r+9/4r-(5r-3/4r+1-3)=0


Domain of the equation: 4r!=0

r!=0/4

r!=0

r∈R



Domain of the equation: 4r+1-3)!=0

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

4r-3)!=-1

r∈R


We add all the numbers together, and all the variables

4r+9/4r-(5r-3/4r-2)=0

We get rid of parentheses

4r+9/4r-5r+3/4r+2=0

We multiply all the terms by the denominator


4r*4r-5r*4r+2*4r+9+3=0

We add all the numbers together, and all the variables

4r*4r-5r*4r+2*4r+12=0

Wy multiply elements

16r^2-20r^2+8r+12=0

We add all the numbers together, and all the variables

-4r^2+8r+12=0

a = -4; b = 8; c = +12;
Δ = b2-4ac
Δ = 82-4·(-4)·12
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*-4}=\frac{-24}{-8}
=+3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*-4}=\frac{8}{-8}
=-1
$