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3x(4x-2)=12
We move all terms to the left:
3x(4x-2)-(12)=0
We multiply parentheses
12x^2-6x-12=0
a = 12; b = -6; c = -12;
Δ = b2-4ac
Δ = -62-4·12·(-12)
Δ = 612
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{17}}{2*12}=\frac{6-6\sqrt{17}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{17}}{2*12}=\frac{6+6\sqrt{17}}{24} $
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