q=270+30q+0.3q^2

Solution for q=270+30q+0.3q^2 equation:

q=270+30q+0.3q^2

We move all terms to the left:

q-(270+30q+0.3q^2)=0

We get rid of parentheses

-0.3q^2-30q+q-270=0

We add all the numbers together, and all the variables

-0.3q^2-29q-270=0

a = -0.3; b = -29; c = -270;
Δ = b2-4ac
Δ = -292-4·(-0.3)·(-270)
Δ = 517
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{-(-29)-\sqrt{517}}{2*-0.3}=\frac{29-\sqrt{517}}{-0.6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{517}}{2*-0.3}=\frac{29+\sqrt{517}}{-0.6}
$