1/8(5y+)=1/4(20+2y)

Solution for 1/8(5y+)=1/4(20+2y) equation:

1/8(5y+)=1/4(20+2y)

We move all terms to the left:

1/8(5y+)-(1/4(20+2y))=0


Domain of the equation: 8(5y+)!=0

y∈R



Domain of the equation: 4(20+2y))!=0

y∈R


We add all the numbers together, and all the variables

1/8(+5y)-(1/4(2y+20))=0

We calculate fractions


(4y2/(8(+5y)*4(2y+20)))+(-8y0/(8(+5y)*4(2y+20)))=0


We calculate terms in parentheses: +(4y2/(8(+5y)*4(2y+20))), so:

4y2/(8(+5y)*4(2y+20))

We multiply all the terms by the denominator


4y2

We add all the numbers together, and all the variables

4y^2

Back to the equation:

+(4y^2)



We calculate terms in parentheses: +(-8y0/(8(+5y)*4(2y+20))), so:

-8y0/(8(+5y)*4(2y+20))

We multiply all the terms by the denominator


-8y0

We add all the numbers together, and all the variables

-8y

Back to the equation:

+(-8y)


We get rid of parentheses

4y^2-8y=0

a = 4; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·4·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*4}=\frac{0}{8}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*4}=\frac{16}{8}
=2
$