3(2x-5)=9(10-x)(2x-5)=9(10-x)

Solution for 3(2x-5)=9(10-x)(2x-5)=9(10-x) equation:

3(2x-5)=9(10-x)(2x-5)=9(10-x)

We move all terms to the left:

3(2x-5)-(9(10-x)(2x-5))=0

We add all the numbers together, and all the variables

3(2x-5)-(9(-1x+10)(2x-5))=0

We multiply parentheses

6x-(9(-1x+10)(2x-5))-15=0

We multiply parentheses ..

-(9(-2x^2+5x+20x-50))+6x-15=0


We calculate terms in parentheses: -(9(-2x^2+5x+20x-50)), so:

9(-2x^2+5x+20x-50)

We multiply parentheses

-18x^2+45x+180x-450

We add all the numbers together, and all the variables

-18x^2+225x-450

Back to the equation:

-(-18x^2+225x-450)


We get rid of parentheses

18x^2-225x+6x+450-15=0

We add all the numbers together, and all the variables

18x^2-219x+435=0

a = 18; b = -219; c = +435;
Δ = b2-4ac
Δ = -2192-4·18·435
Δ = 16641
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16641}=129$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-219)-129}{2*18}=\frac{90}{36}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-219)+129}{2*18}=\frac{348}{36}
=9+2/3
$