65=10(3x+1)+x(9-3x+1)

Solution for 65=10(3x+1)+x(9-3x+1) equation:

65=10(3x+1)+x(9-3x+1)

We move all terms to the left:

65-(10(3x+1)+x(9-3x+1))=0

We add all the numbers together, and all the variables

-(10(3x+1)+x(-3x+10))+65=0


We calculate terms in parentheses: -(10(3x+1)+x(-3x+10)), so:

10(3x+1)+x(-3x+10)

We multiply parentheses

-3x^2+30x+10x+10

We add all the numbers together, and all the variables

-3x^2+40x+10

Back to the equation:

-(-3x^2+40x+10)


We get rid of parentheses

3x^2-40x-10+65=0

We add all the numbers together, and all the variables

3x^2-40x+55=0

a = 3; b = -40; c = +55;
Δ = b2-4ac
Δ = -402-4·3·55
Δ = 940
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{940}=\sqrt{4*235}=\sqrt{4}*\sqrt{235}=2\sqrt{235}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{235}}{2*3}=\frac{40-2\sqrt{235}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{235}}{2*3}=\frac{40+2\sqrt{235}}{6}
$