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0=3x^2+6x-2
We move all terms to the left:
0-(3x^2+6x-2)=0
We add all the numbers together, and all the variables
-(3x^2+6x-2)=0
We get rid of parentheses
-3x^2-6x+2=0
a = -3; b = -6; c = +2;
Δ = b2-4ac
Δ = -62-4·(-3)·2
Δ = 60
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{15}}{2*-3}=\frac{6-2\sqrt{15}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{15}}{2*-3}=\frac{6+2\sqrt{15}}{-6} $
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