1.2q+q^2=0.64

Solution for 1.2q+q^2=0.64 equation:

1.2q+q^2=0.64

We move all terms to the left:

1.2q+q^2-(0.64)=0

We add all the numbers together, and all the variables

q^2+1.2q-0.64=0

a = 1; b = 1.2; c = -0.64;
Δ = b2-4ac
Δ = 1.22-4·1·(-0.64)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1.2)-2}{2*1}=\frac{-3.2}{2}
=-1+1/1.6666666666667
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.2)+2}{2*1}=\frac{0.8}{2}
=0.8/2
$