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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(-2(2q+1)3q), so:

-2(2q+1)3q

We multiply parentheses

-12q^2-6q

Back to the equation:

-(-12q^2-6q)


We get rid of parentheses

12q^2+6q-2q-3=0

We add all the numbers together, and all the variables

12q^2+4q-3=0

a = 12; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·12·(-3)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*12}=\frac{-4-4\sqrt{10}}{24}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*12}=\frac{-4+4\sqrt{10}}{24}
$