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6+12x+3x^2=0
a = 3; b = 12; c = +6;
Δ = b2-4ac
Δ = 122-4·3·6
Δ = 72
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{2}}{2*3}=\frac{-12-6\sqrt{2}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{2}}{2*3}=\frac{-12+6\sqrt{2}}{6} $
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