2/5(25b-40)=1/4(16b+8)

Solution for 2/5(25b-40)=1/4(16b+8) equation:

2/5(25b-40)=1/4(16b+8)

We move all terms to the left:

2/5(25b-40)-(1/4(16b+8))=0


Domain of the equation: 5(25b-40)!=0

b∈R



Domain of the equation: 4(16b+8))!=0

b∈R


We calculate fractions


(8b1/(5(25b-40)*4(16b+8)))+(-5b2/(5(25b-40)*4(16b+8)))=0


We calculate terms in parentheses: +(8b1/(5(25b-40)*4(16b+8))), so:

8b1/(5(25b-40)*4(16b+8))

We multiply all the terms by the denominator


8b1

We add all the numbers together, and all the variables

8b

Back to the equation:

+(8b)



We calculate terms in parentheses: +(-5b2/(5(25b-40)*4(16b+8))), so:

-5b2/(5(25b-40)*4(16b+8))

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 get rid of parentheses

-5b^2+8b=0

a = -5; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-5)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-5}=\frac{-16}{-10}
=1+3/5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-5}=\frac{0}{-10}
=0
$