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1=3x^2
We move all terms to the left:
1-(3x^2)=0
a = -3; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-3)·1
Δ = 12
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*-3}=\frac{0-2\sqrt{3}}{-6} =-\frac{2\sqrt{3}}{-6} =-\frac{\sqrt{3}}{-3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*-3}=\frac{0+2\sqrt{3}}{-6} =\frac{2\sqrt{3}}{-6} =\frac{\sqrt{3}}{-3} $
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