(q+2)(3q-2)=180

Solution for (q+2)(3q-2)=180 equation:

(q+2)(3q-2)=180

We move all terms to the left:

(q+2)(3q-2)-(180)=0

We multiply parentheses ..

(+3q^2-2q+6q-4)-180=0

We get rid of parentheses

3q^2-2q+6q-4-180=0

We add all the numbers together, and all the variables

3q^2+4q-184=0

a = 3; b = 4; c = -184;
Δ = b2-4ac
Δ = 42-4·3·(-184)
Δ = 2224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2224}=\sqrt{16*139}=\sqrt{16}*\sqrt{139}=4\sqrt{139}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{139}}{2*3}=\frac{-4-4\sqrt{139}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{139}}{2*3}=\frac{-4+4\sqrt{139}}{6}
$