6q-q^2=0

Solution for 6q-q^2=0 equation:

6q-q^2=0

We add all the numbers together, and all the variables

-1q^2+6q=0

a = -1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-1)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-1}=\frac{-12}{-2}
=+6
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-1}=\frac{0}{-2}
=0
$