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x2-6x+6=0
We add all the numbers together, and all the variables
x^2-6x+6=0
a = 1; b = -6; c = +6;
Δ = b2-4ac
Δ = -62-4·1·6
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{-(-6)-2\sqrt{3}}{2*1}=\frac{6-2\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{3}}{2*1}=\frac{6+2\sqrt{3}}{2} $
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