4/5(25x+40)-1=-1/2(10x-4)

Solution for 4/5(25x+40)-1=-1/2(10x-4) equation:

4/5(25x+40)-1=-1/2(10x-4)

We move all terms to the left:

4/5(25x+40)-1-(-1/2(10x-4))=0


Domain of the equation: 5(25x+40)!=0

x∈R



Domain of the equation: 2(10x-4))!=0

x∈R


We calculate fractions


(8x1/(5(25x+40)*2(10x-4)))+(-(-5x2)/(5(25x+40)*2(10x-4)))-1=0


We calculate terms in parentheses: +(8x1/(5(25x+40)*2(10x-4))), so:

8x1/(5(25x+40)*2(10x-4))

We multiply all the terms by the denominator


8x1

We add all the numbers together, and all the variables

8x

Back to the equation:

+(8x)



We calculate terms in parentheses: +(-(-5x2)/(5(25x+40)*2(10x-4))), so:

-(-5x2)/(5(25x+40)*2(10x-4))

We add all the numbers together, and all the variables

-(-5x^2)/(5(25x+40)*2(10x-4))

We multiply all the terms by the denominator


-(-5x^2)

We get rid of parentheses

5x^2

Back to the equation:

+(5x^2)


determiningTheFunctionDomain
5x^2+8x-1=0

a = 5; b = 8; c = -1;
Δ = b2-4ac
Δ = 82-4·5·(-1)
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{21}}{2*5}=\frac{-8-2\sqrt{21}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{21}}{2*5}=\frac{-8+2\sqrt{21}}{10}
$