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1/2x^2-12=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
-12*2x^2+1=0
Wy multiply elements
-24x^2+1=0
a = -24; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-24)·1
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-24}=\frac{0-4\sqrt{6}}{-48} =-\frac{4\sqrt{6}}{-48} =-\frac{\sqrt{6}}{-12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-24}=\frac{0+4\sqrt{6}}{-48} =\frac{4\sqrt{6}}{-48} =\frac{\sqrt{6}}{-12} $
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