27+12q=q^2=0

Solution for 27+12q=q^2=0 equation:

27+12q=q^2=0

We move all terms to the left:

27+12q-(q^2)=0

determiningTheFunctionDomain
-q^2+12q+27=0

We add all the numbers together, and all the variables

-1q^2+12q+27=0

a = -1; b = 12; c = +27;
Δ = b2-4ac
Δ = 122-4·(-1)·27
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{7}}{2*-1}=\frac{-12-6\sqrt{7}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{7}}{2*-1}=\frac{-12+6\sqrt{7}}{-2}
$