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2/3x^2=4
We move all terms to the left:
2/3x^2-(4)=0
Domain of the equation: 3x^2!=0We multiply all the terms by the denominator
x^2!=0/3
x^2!=√0
x!=0
x∈R
-4*3x^2+2=0
Wy multiply elements
-12x^2+2=0
a = -12; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-12)·2
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-12}=\frac{0-4\sqrt{6}}{-24} =-\frac{4\sqrt{6}}{-24} =-\frac{\sqrt{6}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-12}=\frac{0+4\sqrt{6}}{-24} =\frac{4\sqrt{6}}{-24} =\frac{\sqrt{6}}{-6} $
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