(3x-9)+(7x^2-3x-3)=x

Solution for (3x-9)+(7x^2-3x-3)=x equation:

(3x-9)+(7x^2-3x-3)=x

We move all terms to the left:

(3x-9)+(7x^2-3x-3)-(x)=0

We add all the numbers together, and all the variables

-1x+(3x-9)+(7x^2-3x-3)=0

We get rid of parentheses

7x^2-1x+3x-3x-9-3=0

We add all the numbers together, and all the variables

7x^2-1x-12=0

a = 7; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·7·(-12)
Δ = 337
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{337}}{2*7}=\frac{1-\sqrt{337}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{337}}{2*7}=\frac{1+\sqrt{337}}{14}
$