1/2(14y+26)=1/5(25y-15)

Solution for 1/2(14y+26)=1/5(25y-15) equation:

1/2(14y+26)=1/5(25y-15)

We move all terms to the left:

1/2(14y+26)-(1/5(25y-15))=0


Domain of the equation: 2(14y+26)!=0

y∈R



Domain of the equation: 5(25y-15))!=0

y∈R


We calculate fractions


(5y2/(2(14y+26)*5(25y-15)))+(-2y1/(2(14y+26)*5(25y-15)))=0


We calculate terms in parentheses: +(5y2/(2(14y+26)*5(25y-15))), so:

5y2/(2(14y+26)*5(25y-15))

We multiply all the terms by the denominator


5y2

We add all the numbers together, and all the variables

5y^2

Back to the equation:

+(5y^2)



We calculate terms in parentheses: +(-2y1/(2(14y+26)*5(25y-15))), so:

-2y1/(2(14y+26)*5(25y-15))

We multiply all the terms by the denominator


-2y1

We add all the numbers together, and all the variables

-2y

Back to the equation:

+(-2y)


We get rid of parentheses

5y^2-2y=0

a = 5; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·5·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*5}=\frac{0}{10}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*5}=\frac{4}{10}
=2/5
$