2r+5=31/2r+2

Solution for 2r+5=31/2r+2 equation:

2r+5=31/2r+2

We move all terms to the left:

2r+5-(31/2r+2)=0


Domain of the equation: 2r+2)!=0

r∈R


We get rid of parentheses

2r-31/2r-2+5=0

We multiply all the terms by the denominator


2r*2r-2*2r+5*2r-31=0

Wy multiply elements

4r^2-4r+10r-31=0

We add all the numbers together, and all the variables

4r^2+6r-31=0

a = 4; b = 6; c = -31;
Δ = b2-4ac
Δ = 62-4·4·(-31)
Δ = 532
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{133}}{2*4}=\frac{-6-2\sqrt{133}}{8}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{133}}{2*4}=\frac{-6+2\sqrt{133}}{8}
$