0.36=q^2+1.6q

Solution for 0.36=q^2+1.6q equation:

0.36=q^2+1.6q

We move all terms to the left:

0.36-(q^2+1.6q)=0

We get rid of parentheses

-q^2-1.6q+0.36=0

We add all the numbers together, and all the variables

-1q^2-1.6q+0.36=0

a = -1; b = -1.6; c = +0.36;
Δ = b2-4ac
Δ = -1.62-4·(-1)·0.36
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{-(-1.6)-2}{2*-1}=\frac{-0.4}{-2}
=0.4/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.6)+2}{2*-1}=\frac{3.6}{-2}
=-1+1/1.25
$