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4/5x^2-12=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
-12*5x^2+4=0
Wy multiply elements
-60x^2+4=0
a = -60; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-60)·4
Δ = 960
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{15}}{2*-60}=\frac{0-8\sqrt{15}}{-120} =-\frac{8\sqrt{15}}{-120} =-\frac{\sqrt{15}}{-15} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{15}}{2*-60}=\frac{0+8\sqrt{15}}{-120} =\frac{8\sqrt{15}}{-120} =\frac{\sqrt{15}}{-15} $
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