1/2(14b+2)=1/5(25b+15)

Solution for 1/2(14b+2)=1/5(25b+15) equation:

1/2(14b+2)=1/5(25b+15)

We move all terms to the left:

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


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

b∈R



Domain of the equation: 5(25b+15))!=0

b∈R


We calculate fractions


(5b2/(2(14b+2)*5(25b+15)))+(-2b1/(2(14b+2)*5(25b+15)))=0


We calculate terms in parentheses: +(5b2/(2(14b+2)*5(25b+15))), so:

5b2/(2(14b+2)*5(25b+15))

We multiply all the terms by the denominator


5b2

We add all the numbers together, and all the variables

5b^2

Back to the equation:

+(5b^2)



We calculate terms in parentheses: +(-2b1/(2(14b+2)*5(25b+15))), so:

-2b1/(2(14b+2)*5(25b+15))

We multiply all the terms by the denominator


-2b1

We add all the numbers together, and all the variables

-2b

Back to the equation:

+(-2b)


We get rid of parentheses

5b^2-2b=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:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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