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12x^2+4x=6
We move all terms to the left:
12x^2+4x-(6)=0
a = 12; b = 4; c = -6;
Δ = b2-4ac
Δ = 42-4·12·(-6)
Δ = 304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{19}}{2*12}=\frac{-4-4\sqrt{19}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{19}}{2*12}=\frac{-4+4\sqrt{19}}{24} $
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