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2x^2-12x=6=0
We move all terms to the left:
2x^2-12x-(6)=0
a = 2; b = -12; c = -6;
Δ = b2-4ac
Δ = -122-4·2·(-6)
Δ = 192
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{3}}{2*2}=\frac{12-8\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{3}}{2*2}=\frac{12+8\sqrt{3}}{4} $
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