1/2(16x+20)=1/5(25x-40)

Solution for 1/2(16x+20)=1/5(25x-40) equation:

1/2(16x+20)=1/5(25x-40)

We move all terms to the left:

1/2(16x+20)-(1/5(25x-40))=0


Domain of the equation: 2(16x+20)!=0

x∈R



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

x∈R


We calculate fractions


(5x2/(2(16x+20)*5(25x-40)))+(-2x1/(2(16x+20)*5(25x-40)))=0


We calculate terms in parentheses: +(5x2/(2(16x+20)*5(25x-40))), so:

5x2/(2(16x+20)*5(25x-40))

We multiply all the terms by the denominator


5x2

We add all the numbers together, and all the variables

5x^2

Back to the equation:

+(5x^2)



We calculate terms in parentheses: +(-2x1/(2(16x+20)*5(25x-40))), so:

-2x1/(2(16x+20)*5(25x-40))

We multiply all the terms by the denominator


-2x1

We add all the numbers together, and all the variables

-2x

Back to the equation:

+(-2x)


We get rid of parentheses

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

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