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6x=1-3x^2
We move all terms to the left:
6x-(1-3x^2)=0
We get rid of parentheses
3x^2+6x-1=0
a = 3; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·3·(-1)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*3}=\frac{-6-4\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*3}=\frac{-6+4\sqrt{3}}{6} $
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