15q-3q(4q-6)=3q+9-9q

Solution for 15q-3q(4q-6)=3q+9-9q equation:

15q-3q(4q-6)=3q+9-9q

We move all terms to the left:

15q-3q(4q-6)-(3q+9-9q)=0

We add all the numbers together, and all the variables

15q-3q(4q-6)-(-6q+9)=0

We multiply parentheses

-12q^2+15q+18q-(-6q+9)=0

We get rid of parentheses

-12q^2+15q+18q+6q-9=0

We add all the numbers together, and all the variables

-12q^2+39q-9=0

a = -12; b = 39; c = -9;
Δ = b2-4ac
Δ = 392-4·(-12)·(-9)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-33}{2*-12}=\frac{-72}{-24}
=+3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+33}{2*-12}=\frac{-6}{-24}
=1/4
$