4/5(20x-10)=3/4(16-23x)

Solution for 4/5(20x-10)=3/4(16-23x) equation:

4/5(20x-10)=3/4(16-23x)

We move all terms to the left:

4/5(20x-10)-(3/4(16-23x))=0


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

x∈R



Domain of the equation: 4(16-23x))!=0

x∈R


We add all the numbers together, and all the variables

4/5(20x-10)-(3/4(-23x+16))=0

We calculate fractions


(16x(-)/(5(20x-10)*4(-23x+16)))+(-15x2/(5(20x-10)*4(-23x+16)))=0


We calculate terms in parentheses: +(16x(-)/(5(20x-10)*4(-23x+16))), so:

16x(-)/(5(20x-10)*4(-23x+16))

We add all the numbers together, and all the variables

16x0/(5(20x-10)*4(-23x+16))

We multiply all the terms by the denominator


16x0

We add all the numbers together, and all the variables

16x

Back to the equation:

+(16x)



We calculate terms in parentheses: +(-15x2/(5(20x-10)*4(-23x+16))), so:

-15x2/(5(20x-10)*4(-23x+16))

We multiply all the terms by the denominator


-15x2

We add all the numbers together, and all the variables

-15x^2

Back to the equation:

+(-15x^2)


We get rid of parentheses

-15x^2+16x=0

a = -15; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-15)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-15}=\frac{-32}{-30}
=1+1/15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-15}=\frac{0}{-30}
=0
$