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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3x^2-1x+30x+10)+9x-65=0

We get rid of parentheses

-3x^2-1x+30x+9x+10-65=0

We add all the numbers together, and all the variables

-3x^2+38x-55=0

a = -3; b = 38; c = -55;
Δ = b2-4ac
Δ = 382-4·(-3)·(-55)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-28}{2*-3}=\frac{-66}{-6}
=+11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+28}{2*-3}=\frac{-10}{-6}
=1+2/3
$