20s^2-32=0

Solution for 20s^2-32=0 equation:

20s^2-32=0

a = 20; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·20·(-32)
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{10}}{2*20}=\frac{0-16\sqrt{10}}{40}
=-\frac{16\sqrt{10}}{40}
=-\frac{2\sqrt{10}}{5}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{10}}{2*20}=\frac{0+16\sqrt{10}}{40}
=\frac{16\sqrt{10}}{40}
=\frac{2\sqrt{10}}{5}
$