-3q+9=3q^2-6q+3

Solution for -3q+9=3q^2-6q+3 equation:

-3q+9=3q^2-6q+3

We move all terms to the left:

-3q+9-(3q^2-6q+3)=0

We get rid of parentheses

-3q^2-3q+6q-3+9=0

We add all the numbers together, and all the variables

-3q^2+3q+6=0

a = -3; b = 3; c = +6;
Δ = b2-4ac
Δ = 32-4·(-3)·6
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*-3}=\frac{-12}{-6}
=+2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*-3}=\frac{6}{-6}
=-1
$