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1/5x^2-3=5
We move all terms to the left:
1/5x^2-3-(5)=0
Domain of the equation: 5x^2!=0We add all the numbers together, and all the variables
x^2!=0/5
x^2!=√0
x!=0
x∈R
1/5x^2-8=0
We multiply all the terms by the denominator
-8*5x^2+1=0
Wy multiply elements
-40x^2+1=0
a = -40; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-40)·1
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-40}=\frac{0-4\sqrt{10}}{-80} =-\frac{4\sqrt{10}}{-80} =-\frac{\sqrt{10}}{-20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-40}=\frac{0+4\sqrt{10}}{-80} =\frac{4\sqrt{10}}{-80} =\frac{\sqrt{10}}{-20} $
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