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4x(6x-3)=4
We move all terms to the left:
4x(6x-3)-(4)=0
We multiply parentheses
24x^2-12x-4=0
a = 24; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·24·(-4)
Δ = 528
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{33}}{2*24}=\frac{12-4\sqrt{33}}{48} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{33}}{2*24}=\frac{12+4\sqrt{33}}{48} $
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