4s^2-32=0

Solution for 4s^2-32=0 equation:

4s^2-32=0

a = 4; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·4·(-32)
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*4}=\frac{0-16\sqrt{2}}{8}
=-\frac{16\sqrt{2}}{8}
=-2\sqrt{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*4}=\frac{0+16\sqrt{2}}{8}
=\frac{16\sqrt{2}}{8}
=2\sqrt{2}
$