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4/5x^2-2=0
Domain of the equation: 5x^2!=0We multiply all the terms by the denominator
x^2!=0/5
x^2!=√0
x!=0
x∈R
-2*5x^2+4=0
Wy multiply elements
-10x^2+4=0
a = -10; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-10)·4
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*-10}=\frac{0-4\sqrt{10}}{-20} =-\frac{4\sqrt{10}}{-20} =-\frac{\sqrt{10}}{-5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-10}=\frac{0+4\sqrt{10}}{-20} =\frac{4\sqrt{10}}{-20} =\frac{\sqrt{10}}{-5} $
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