0=3x^2-18x-6

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Solution for 0=3x^2-18x-6 equation:



0=3x^2-18x-6
We move all terms to the left:
0-(3x^2-18x-6)=0
We add all the numbers together, and all the variables
-(3x^2-18x-6)=0
We get rid of parentheses
-3x^2+18x+6=0
a = -3; b = 18; c = +6;
Δ = b2-4ac
Δ = 182-4·(-3)·6
Δ = 396
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:
$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{11}}{2*-3}=\frac{-18-6\sqrt{11}}{-6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{11}}{2*-3}=\frac{-18+6\sqrt{11}}{-6} $

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