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2/3x^2+4=16
We move all terms to the left:
2/3x^2+4-(16)=0
Domain of the equation: 3x^2!=0We add all the numbers together, and all the variables
x^2!=0/3
x^2!=√0
x!=0
x∈R
2/3x^2-12=0
We multiply all the terms by the denominator
-12*3x^2+2=0
Wy multiply elements
-36x^2+2=0
a = -36; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-36)·2
Δ = 288
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*-36}=\frac{0-12\sqrt{2}}{-72} =-\frac{12\sqrt{2}}{-72} =-\frac{\sqrt{2}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*-36}=\frac{0+12\sqrt{2}}{-72} =\frac{12\sqrt{2}}{-72} =\frac{\sqrt{2}}{-6} $
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