q2-256=0

Solution for q2-256=0 equation:

q2-256=0

We add all the numbers together, and all the variables

q^2-256=0

a = 1; b = 0; c = -256;
Δ = b2-4ac
Δ = 02-4·1·(-256)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*1}=\frac{-32}{2}
=-16
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*1}=\frac{32}{2}
=16
$