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6x^2-3=12x
We move all terms to the left:
6x^2-3-(12x)=0
a = 6; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·6·(-3)
Δ = 216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{6}}{2*6}=\frac{12-6\sqrt{6}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{6}}{2*6}=\frac{12+6\sqrt{6}}{12} $
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