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1/2x^2=256
We move all terms to the left:
1/2x^2-(256)=0
Domain of the equation: 2x^2!=0We multiply all the terms by the denominator
x^2!=0/2
x^2!=√0
x!=0
x∈R
-256*2x^2+1=0
Wy multiply elements
-512x^2+1=0
a = -512; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-512)·1
Δ = 2048
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{2}}{2*-512}=\frac{0-32\sqrt{2}}{-1024} =-\frac{32\sqrt{2}}{-1024} =-\frac{\sqrt{2}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{2}}{2*-512}=\frac{0+32\sqrt{2}}{-1024} =\frac{32\sqrt{2}}{-1024} =\frac{\sqrt{2}}{-32} $
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