0.5x^2-4=0

Solution for 0.5x^2-4=0 equation:

0.5x^2-4=0

a = 0.5; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·0.5·(-4)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*0.5}=\frac{0-2\sqrt{2}}{1}
=-\frac{2\sqrt{2}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*0.5}=\frac{0+2\sqrt{2}}{1}
=\frac{2\sqrt{2}}{1}
$