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6q^2-4q=8q+12
We move all terms to the left:
6q^2-4q-(8q+12)=0
We get rid of parentheses
6q^2-4q-8q-12=0
We add all the numbers together, and all the variables
6q^2-12q-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:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{3}}{2*6}=\frac{12-12\sqrt{3}}{12} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{3}}{2*6}=\frac{12+12\sqrt{3}}{12} $
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