8/9(-18x+9)=4/5(-10+20x)

Solution for 8/9(-18x+9)=4/5(-10+20x) equation:

8/9(-18x+9)=4/5(-10+20x)

We move all terms to the left:

8/9(-18x+9)-(4/5(-10+20x))=0


Domain of the equation: 9(-18x+9)!=0

x∈R



Domain of the equation: 5(-10+20x))!=0

x∈R


We add all the numbers together, and all the variables

8/9(-18x+9)-(4/5(20x-10))=0

We calculate fractions


(40x2/(9(-18x+9)*5(20x-10)))+(-36x0/(9(-18x+9)*5(20x-10)))=0


We calculate terms in parentheses: +(40x2/(9(-18x+9)*5(20x-10))), so:

40x2/(9(-18x+9)*5(20x-10))

We multiply all the terms by the denominator


40x2

We add all the numbers together, and all the variables

40x^2

Back to the equation:

+(40x^2)



We calculate terms in parentheses: +(-36x0/(9(-18x+9)*5(20x-10))), so:

-36x0/(9(-18x+9)*5(20x-10))

We multiply all the terms by the denominator


-36x0

We add all the numbers together, and all the variables

-36x

Back to the equation:

+(-36x)


We get rid of parentheses

40x^2-36x=0

a = 40; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·40·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*40}=\frac{0}{80}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*40}=\frac{72}{80}
=9/10
$