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1/2x^2+2=52
We move all terms to the left:
1/2x^2+2-(52)=0
Domain of the equation: 2x^2!=0We add all the numbers together, and all the variables
x^2!=0/2
x^2!=√0
x!=0
x∈R
1/2x^2-50=0
We multiply all the terms by the denominator
-50*2x^2+1=0
Wy multiply elements
-100x^2+1=0
a = -100; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-100)·1
Δ = 400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*-100}=\frac{-20}{-200} =1/10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*-100}=\frac{20}{-200} =-1/10 $
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