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3x^2=12x-4
We move all terms to the left:
3x^2-(12x-4)=0
We get rid of parentheses
3x^2-12x+4=0
a = 3; b = -12; c = +4;
Δ = b2-4ac
Δ = -122-4·3·4
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{6}}{2*3}=\frac{12-4\sqrt{6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{6}}{2*3}=\frac{12+4\sqrt{6}}{6} $
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