(3q-1)(q+2)=-3

Solution for (3q-1)(q+2)=-3 equation:

(3q-1)(q+2)=-3

We move all terms to the left:

(3q-1)(q+2)-(-3)=0

We add all the numbers together, and all the variables

(3q-1)(q+2)+3=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3q^2+5q+1=0

a = 3; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·3·1
Δ = 13
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{-(5)-\sqrt{13}}{2*3}=\frac{-5-\sqrt{13}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{13}}{2*3}=\frac{-5+\sqrt{13}}{6}
$