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2(4c+1)c=12
We move all terms to the left:
2(4c+1)c-(12)=0
We multiply parentheses
8c^2+2c-12=0
a = 8; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·8·(-12)
Δ = 388
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{97}}{2*8}=\frac{-2-2\sqrt{97}}{16} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{97}}{2*8}=\frac{-2+2\sqrt{97}}{16} $
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