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1/2x^2-5=5
We move all terms to the left:
1/2x^2-5-(5)=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^2-10=0
We multiply all the terms by the denominator
-10*2x^2+1=0
Wy multiply elements
-20x^2+1=0
a = -20; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-20)·1
Δ = 80
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-20}=\frac{0-4\sqrt{5}}{-40} =-\frac{4\sqrt{5}}{-40} =-\frac{\sqrt{5}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-20}=\frac{0+4\sqrt{5}}{-40} =\frac{4\sqrt{5}}{-40} =\frac{\sqrt{5}}{-10} $
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