11/22r-3-5/12r=7

Solution for 11/22r-3-5/12r=7 equation:

11/22r-3-5/12r=7

We move all terms to the left:

11/22r-3-5/12r-(7)=0


Domain of the equation: 22r!=0

r!=0/22

r!=0

r∈R



Domain of the equation: 12r!=0

r!=0/12

r!=0

r∈R


We add all the numbers together, and all the variables

11/22r-5/12r-10=0

We calculate fractions


132r/264r^2+(-110r)/264r^2-10=0

We multiply all the terms by the denominator


132r+(-110r)-10*264r^2=0

Wy multiply elements

-2640r^2+132r+(-110r)=0

We get rid of parentheses

-2640r^2+132r-110r=0

We add all the numbers together, and all the variables

-2640r^2+22r=0

a = -2640; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-2640)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-2640}=\frac{-44}{-5280}
=1/120
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-2640}=\frac{0}{-5280}
=0
$