3q2-q-6=-3q2

Solution for 3q2-q-6=-3q2 equation:

3q^2-q-6=-3q^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

3q^2-(-3q^2)-1q-6=0

We get rid of parentheses

3q^2+3q^2-1q-6=0

We add all the numbers together, and all the variables

6q^2-1q-6=0

a = 6; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·6·(-6)
Δ = 145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{145}}{2*6}=\frac{1-\sqrt{145}}{12}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{145}}{2*6}=\frac{1+\sqrt{145}}{12}
$