1/2(2x+9)=1/5(10x-5)

Solution for 1/2(2x+9)=1/5(10x-5) equation:

1/2(2x+9)=1/5(10x-5)

We move all terms to the left:

1/2(2x+9)-(1/5(10x-5))=0


Domain of the equation: 2(2x+9)!=0

x∈R



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

x∈R


We calculate fractions


(5x1/(2(2x+9)*5(10x-5)))+(-2x2/(2(2x+9)*5(10x-5)))=0


We calculate terms in parentheses: +(5x1/(2(2x+9)*5(10x-5))), so:

5x1/(2(2x+9)*5(10x-5))

We multiply all the terms by the denominator


5x1

We add all the numbers together, and all the variables

5x

Back to the equation:

+(5x)



We calculate terms in parentheses: +(-2x2/(2(2x+9)*5(10x-5))), so:

-2x2/(2(2x+9)*5(10x-5))

We multiply all the terms by the denominator


-2x2

We add all the numbers together, and all the variables

-2x^2

Back to the equation:

+(-2x^2)


We get rid of parentheses

-2x^2+5x=0

a = -2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-2)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-2}=\frac{-10}{-4}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-2}=\frac{0}{-4}
=0
$