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6a^2+12a-12=0
a = 6; b = 12; c = -12;
Δ = b2-4ac
Δ = 122-4·6·(-12)
Δ = 432
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{3}}{2*6}=\frac{-12-12\sqrt{3}}{12} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{3}}{2*6}=\frac{-12+12\sqrt{3}}{12} $
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