32=q2

Solution for 32=q2 equation:

32=q2

We move all terms to the left:

32-(q2)=0

We add all the numbers together, and all the variables

-1q^2+32=0

a = -1; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-1)·32
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-1}=\frac{0-8\sqrt{2}}{-2}
=-\frac{8\sqrt{2}}{-2}
=-\frac{4\sqrt{2}}{-1}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-1}=\frac{0+8\sqrt{2}}{-2}
=\frac{8\sqrt{2}}{-2}
=\frac{4\sqrt{2}}{-1}
$