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3x(4x-3)=12
We move all terms to the left:
3x(4x-3)-(12)=0
We multiply parentheses
12x^2-9x-12=0
a = 12; b = -9; c = -12;
Δ = b2-4ac
Δ = -92-4·12·(-12)
Δ = 657
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{73}}{2*12}=\frac{9-3\sqrt{73}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{73}}{2*12}=\frac{9+3\sqrt{73}}{24} $
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