6q=0.1q^2+50

Solution for 6q=0.1q^2+50 equation:

6q=0.1q^2+50

We move all terms to the left:

6q-(0.1q^2+50)=0

We get rid of parentheses

-0.1q^2+6q-50=0

a = -0.1; b = 6; c = -50;
Δ = b2-4ac
Δ = 62-4·(-0.1)·(-50)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*-0.1}=\frac{-10}{-0.2}
=+50
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*-0.1}=\frac{-2}{-0.2}
=+10
$