4q^2-8=0

Solution for 4q^2-8=0 equation:

4q^2-8=0

a = 4; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·4·(-8)
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*4}=\frac{0-8\sqrt{2}}{8}
=-\frac{8\sqrt{2}}{8}
=-\sqrt{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*4}=\frac{0+8\sqrt{2}}{8}
=\frac{8\sqrt{2}}{8}
=\sqrt{2}
$