3/4(2r-4)=1/5(36-r)

Solution for 3/4(2r-4)=1/5(36-r) equation:

3/4(2r-4)=1/5(36-r)

We move all terms to the left:

3/4(2r-4)-(1/5(36-r))=0


Domain of the equation: 4(2r-4)!=0

r∈R



Domain of the equation: 5(36-r))!=0

r∈R


We add all the numbers together, and all the variables

3/4(2r-4)-(1/5(-1r+36))=0

We calculate fractions


(15r(-)/(4(2r-4)*5(-1r+36)))+(-4r2/(4(2r-4)*5(-1r+36)))=0


We calculate terms in parentheses: +(15r(-)/(4(2r-4)*5(-1r+36))), so:

15r(-)/(4(2r-4)*5(-1r+36))

We add all the numbers together, and all the variables

15r0/(4(2r-4)*5(-1r+36))

We multiply all the terms by the denominator


15r0

We add all the numbers together, and all the variables

15r

Back to the equation:

+(15r)



We calculate terms in parentheses: +(-4r2/(4(2r-4)*5(-1r+36))), so:

-4r2/(4(2r-4)*5(-1r+36))

We multiply all the terms by the denominator


-4r2

We add all the numbers together, and all the variables

-4r^2

Back to the equation:

+(-4r^2)


We get rid of parentheses

-4r^2+15r=0

a = -4; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·(-4)·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*-4}=\frac{-30}{-8}
=3+3/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*-4}=\frac{0}{-8}
=0
$