31/2r+2=2r+5

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

31/2r+2=2r+5

We move all terms to the left:

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


Domain of the equation: 2r!=0

r!=0/2

r!=0

r∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-4r^2-10r+4r+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}
$