-(2x)^2=-4

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Solution for -(2x)^2=-4 equation:



-(2x)^2=-4
We move all terms to the left:
-(2x)^2-(-4)=0
We add all the numbers together, and all the variables
-2x^2+4=0
a = -2; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-2)·4
Δ = 32
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:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-2}=\frac{0-4\sqrt{2}}{-4} =-\frac{4\sqrt{2}}{-4} =-\frac{\sqrt{2}}{-1} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-2}=\frac{0+4\sqrt{2}}{-4} =\frac{4\sqrt{2}}{-4} =\frac{\sqrt{2}}{-1} $

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